Mathematics-20.Unit-16 Circle

1. 1 STANDARD EQUATION OF A CIRCLE 3 EQUATION OF CIRCLE IN DIFFERENT CONDITIONS 3 CONDITION FOR THE GENERAL EQUATION OF SECOND DEGREE IN X AND Y TO REPRESENT A CIRCLE 4 GENERAL EQUATION OF A CIRCLE 4 EQUATION OF A CIRCLE WHOSE END POINTS OF ANY DIAMETER IS GIVEN 5 INTERCEPT MADE BY THE CIRCLE ON THE AXIS 5 PARAMETRIC EQUATION OF A CIRCLE 7 POSITION OF A POINT WITH RESPECT TO A CIRCLE 7 INTERSECTION OF A LINE WITH A CIRCLE 7 EQUATION OF CHORD WHOSE MID POINT IS GIVEN 8 EQUATION OF TANGENTS 8 EQUATION OF TANGENT FROM ANY POINT OUTSIDE THE CIRCLE 9 EQUATION OF NORMAL 10 LENGTH OF THE TANGENT 10 INTERSECTION OF TWO CIRCLES 10 ANGLE OF INTERSECTION OF TWO CIRCLES 12 ORTHOGONAL INTERSECTION OF TWO CIRCLES 13 2. 12 2.1 13 2.2 14 2.3 15 3. 16 SYLLABUS Equation of a circle in various forms, equation of tangent, normal and chord. Parametric equations of a circle, intersection of a circle with a straight line or a circle, equation of a circle through the points of intersection of two circles and those of a circle and a straight line. 1. DEFINITION A circle is the locus of a point which moves in a plane such that its distance from a fixed point is always constant. The fixed point is called the centre of the circle and the constant distance, the radius of the circle. STANDARD EQUATION OF A CIRCLE The equation of a circle with the centre at (a, b) and radius r, is given by (x – a)2 + (y – b)2 = r2 .If the centre of the circle is at the origin and the radius is r, then equation of circle is x2 + y2 = r2. EQUATION OF CIRCLE IN DIFFERENT CONDITIONS Condition Equation (i) Touches both the axes with centre (a, a) and radius a (x–a)2 + (y–a)2 = a2 (ii) Touches x–axis only with centre (α, a) and radius a (x–α)2 + (y–a)2 = a2 (iii) Touches y–axis only with centre (a, β) and radius a (x–a)2 + (y–β)2 = a2 (iv) Passes through the origin with centre and radius . x2 + y2 – αx – βy = 0 CONDITION FOR THE GENERAL EQUATION OF SECOND DEGREE IN X AND Y TO REPRESENT A CIRCLE The general equation of second degree ax2+2hxy+by2+2gx+2fy+c=0 represent a circle ,if ⮚ Coefficient of x2=coefficient of y2 i.e. a=b ⮚ Coefficient of xy=0 i.e. h=0. GENERAL EQUATION OF A CIRCLE The general equation of a circle is of the form x2 + y2 + 2gx + 2fy + c = 0 where g, f and c are constants. The given equation of circle in the standard form can be written as . Hence the coordinates of its centre are (-g, -f) and radius is . Case I: if >0, a real circle is possible. Case II: if =0 , the circle is called a point circle. Case III: if < 0 no real circle is possible. Working rule to find the centre and radius of a circle whose equation is given: STEP I: Make The coefficients of x2 and y2 equal to 1 and right hand side equal to zero. STEP II: The coordinate of the centre will be (a, b) where a= (coefficient of x) and b= (coefficient of y). STEP III: Radius= Illustration 1: Find the centre and radius of the circle (A) 2x2+2y2-4x-5y-1=0. (B) for some . Solution: (A) The given equation of circle can be written as x2+y2-2x-5/2y-1/2=0 g=-1, f=-5/4 c=-1/2 Hence the centre is (1,5/4) and radius is . (B) The given equation can be written as Since there is no term of xy in the equation of circle Hence the given equation reduces to x2+y2-x+2y-2=0. Centre is (1/2, -1) and radius is EQUATION OF A CIRCLE WHOSE END POINTS OF ANY DIAMETER IS GIVEN Equation of the circle with points P(x1, y1) and Q(x2, y2) as extremities of a diameter is given by (x – x1)(x – x2) + (y – y1)(y – y2) = 0. INTERCEPT MADE BY THE CIRCLE ON THE AXIS Let the equation of circle be x2 + y2 + 2gx + 2fy + c = 0………(1) X- INTERCEPT: Intercept made by the circle on the x-axis is called the X-intercept. The circle will intersect the x-axis where y = 0 x2 + 2gx + c = 0 …(2) Here three cases arises Case I: If discriminant > 0 i.e. >0 circle will intersect the axis at two distinct and real points let A (x1, 0) and B (x2, 0). Length of the intercept . Case II: If discriminant = 0 i.e. g2=c. Then the circle will touch the x-axis. In this case length of the intercept made by the circle on the x-axis will be zero. Case III: If discriminant < 0 i.e. <0, in this case circle will neither touch nor intersect the x-axis Y- INTERCEPT: Intercept made by the circle on the y-axis is called the Y-intercept. The circle will intersect the y-axis where x = 0 y2 + 2fy + c = 0 …(2) Again three cases arises Case I: When discriminant > 0 i.e. >0 circle will intersect the y-axis at two distinct and real points say A(0, y1) and B(0, y2). Length of the y-intercept Case II: If discriminant = 0 i.e. =0 Then the circle will touch the axis. In this case length of the intercept made by the circle on the x-axis will be zero. Case III: If discriminant < 0, i.e. <0 in this case circle will neither touch nor intersect the axes. Illustration 2: Find equation of the circle touching y–axis at (0, 3) and making intercept of 8 units on the x–axis. Solution: Let the equation of circle be x2 + y2 + 2gx + 2fy + c = 0 putting x = 0, we get y2 + 2fy + c = 0 … (1) As it touches y−axis at (0, 3), (1) must be of the form (y − 3)2 = 0 ⇒ y2 − 6x + 9 = 0 Comparing, we get, f = − 3 and c = 9 Now, putting y = 0, we get x2 + 2gx + c = 0 So, |x1 − x2| = = 2 = 8 ⇒ g2 − c = 16 ⇒ g2 = 25 ⇒ g = ± 5 so, the equation is x2 + y2 ± 10x − 6y + 9 = 0 PARAMETRIC EQUATION OF A CIRCLE Let the equation of circle be (x – a)2 + (y – b)2 = r2. Hence from the diagram it is clear that co-ordinates of any point on the circle can be taken as (a + r cosθ, b + r sinθ) where 0 ≤ θ < 2π . x = a + r cosθ and y = b + r sinθ is called the parameter equation of the circle. POSITION OF A POINT WITH RESPECT TO A CIRCLE Let the equation of the given circle be x2+y2+2gx+2fy+c=0 and the given point be P(α,β). Now, If the distance of the point P(α,β) from the centre of the circle(O) is greater than the radius of the circle i.e. , then the point will lie outside the circle. α2 + β2 + 2αg + 2βf + c > 0, point P will lie outside the circle. Similarly If α2 + β2 + 2αg + 2βf + c = 0, point P will lie outside the circle. If α2 + β2 + 2αg + 2βf + c <0 0, point P will lie outside the circle. INTERSECTION OF A LINE WITH A CIRCLE CONCEPT 1: Let S=0 be a circle with centre C and radius r ,L=0 be a line and p be perpendicular distance from the centre to the line L=0. Case I: If p>r i.e. the distance from the centre to the line is greater than the radius then the circle doesn’t intersect the line. Case II: If p=r i.e. the distance from the centre to the line is equal to the radius then the circle will intersect the line in one and only one point i.e. circle touches the line. This is also called the condition of tangency Case III: If p r1 +r2. Then both the circles will be non intersecting. In this case there exist four common tangents to these two circles. Two direct common tangents and two transverse common tangent. Working Rule to find direct common tangent: Step I: First find the point of intersection of direct common tangents say Q, which divides O1 O2 externally in r1 : r2 Step II: Write the equation of any line passing through Q (α, β), i.e. y-β = m (x-α)…….(1) Step III: Find the two values of m, using the fact that the length of the perpendicular on (1) from the centre of one circle is equal to its radius. Step IV: Substitutes these values of ‘m’ in (1), the equation of the two direct common tangents can be obtained. Working Rule to find transverse common tangent: To fine the equations of transverse common tangent first find the point of intersection of transverse common tangents say P, which divides O1O2 internally in r1:r2. Then follow the step 2, 3 and 4. Case II: If the distance between the centres of the given circle is equal to sum of theirs radii. In this case both the circle will be touching each other externally. In this case two direct common tangents are real and distinct while the transverse tangents are coincident. O1 O2 =|r1+r2| The point of contact P can be find by using the fact that it divides O1 O2 internally in r1 : r2 . Case III: It the distance between the centres of the given circles is equal to difference of their radii i.e. |O1 O2| = |r1-r2|, both the circles touches each other internally. In this case point of contact divides O1 O2 externally in r1 : r2. In this case only one common tangent exist. Case IV: It the distance between the centres of two given circle is less then the sum of their radii but greater then the difference of their radii i.e. |r1 - r2| < O1 O2 < r1 + r2, in this case both the circle will intersect at two real and distinct points. In this case there exist two direct common tangents. ANGLE OF INTERSECTION OF TWO CIRCLES Angle of intersection of two circles is defined as the angle between the tangents at this point of intersection. Here angle ∠O1PO2 = π - θ ORTHOGONAL INTERSECTION OF TWO CIRCLES The two circles are said to intersect orthogonally if the angle between the tangents at their point of intersection is 900. The condition for two circles S1=O and S2=O to cut each other orthogonally is Note: It two circle are intersecting orthogonally the tangent to one circle at the point of the intersection passes through the centre of other circle. Case V: It the distance between the centres is less than the difference of their radii, i.e. |O1 O2| < |r1-r2|, in this case one circle will lie completely inside the other circle. Hence there will be no common tangent. Illustration 6: If the circles x2 + y2 + 2ax + 2by = 0 and x2 + y2 + 2bx + 2cy = 0 touch each other, then show that b2=ac. Solution I : Both the circles are touching each other. Hence distance between the centres should be equal to sum of radii. Centre are (–a, –b) and (–b, –c) and their radii are respectively squaring both sides ⇒ b4 – 2acb2 + a2c2 = 0 ⇒ (b2 – ac)2 = 0 ⇒ b2 = ac Solution II : Clearly both the circles are passing through the origin. Hence they will be ouching each other at (0, 0). Hence at (0, 0) they will be having a common tangent. ⇒ 0.x + 0.y + a(x + 0) + b(y + 0) = 0 ⇒ ax + by = 0 …(1) ⇒ 0.x + 0.y + b(x + 0) + c(y + 0) = 0 ⇒ bx + cy = 0 …(2) (1) and (2) identical a/b = b/c ⇒ b2 = ac Illustration 7: Find the locus of centre of a circle that passes through (a, b) and cuts the circle x2+y2=a2 orthogonally. Solution I : Let the circle be x2 + y2 + 2gx + 2fy + c = 0. It passes through (a, b) thus a2 + b2 + 2ag + 2fb + c = 0. It also cuts x2 + y2 = a2 orthogonally, thus 2g.0 + 2f.0 = c – a2 2ag + 2fb + b2 + 2a2 = 0. Thus locus of centre is Solution II : Let the centre of the given circle be P(h, k). Now this circle is intersecting the x2 + y2 =a2 orthogonallyPQR will be a right angle triangle. Hence PR2 = PQ2 + QR2 , h2 + k2 = (h – a)2 + (k – b)2 + a2 ⇒ 2ah + 2bk = 2a2 + b2 Hence the locus is 2ax + 2by = 2a2 + b2 Illustration 8: Find the radius of bigger circle touching the circle x2+y2-4x-4y+4=0 Solution: Let the centre of required circle be (h, h). We have , CB = h + 2, BD = h – 2 2. LOCATION OF A CIRCLE IN RELATION TO A CIRCLE Let S1 ≡ x2 + y2 + 2g1x + 2f1y+ c1 = 0 and S2 ≡ x2 + y2 + 2g2x + 2f2y+ c1 = 0 be two circles. Let D be the discriminant for the quadratic equation in x (or y) obtained by eliminating y (or x) from the two equations of the circle. Then (i) they are two intersecting circles if D > 0 (ii) they are nonintersecting (no common points) if D < 0 (iii) they touch each other if D = 0 (iv) If D < 0, i.e., the circles are nonintersecting then (a) S1 = 0 is outside S2 = 0 if S2 (-g1, –f1) > 0 or S1 (–g2, –f2) > 0; equivalently, AB > r1 + r2 where A, B are centres and r1, r2 are radii respectively. (b) S1 = 0 is inside S2 = 0 if S2 (–g1, –f1) < 0; equivalently, AB < |r2 – r1| (v) If D = 0, i.e., then the circles touches each other (c) externally if AB = r1 + r2 (d) internally if AB = | r1 – r2| 2.1 CHORD OF CONTACT From a point P(x1, y1) out side the circle two tangents PA and PB can be drawn to the circle. The chord AB joining the points of contact A and B of the tangents from P is called the chord of contact of P(x1, y1) with respect to the circle. Its equation is given by T = 0. Illustration 9: If the chord of contact of the tangents drawn to x2+y2=a2 from any point on x2+y2=b2, touches the circle x2+y2=c2, then show that a2=bc Solution 1: Let P(x1, y1) be any point on x2+y2=b2 i.e. x12+y12=b2. Equation of corresponding chord of contact is xx1+yy1-a2=0. It touches x2+y2=c2 Solution 2: In OAC …(1) and in OAP …(2) from (1) and (2) a2 = bc Illustration 10: Tangents to the circle x2 + y2 = a2 cut the circle x2 + y2 = 2a2 at P and Q. Prove that tangents at P and Q to the circle x2 + y2 = 2a2 intersect at right angles. Solution: Equation of tangent at any point (a cosθ, a sin θ) to the circle x2 + y2 = a2 is xcosθ + ysin θ = a ………(1) Let the point of intersection of the tangents at P and Q be (h, k). If the tangents at P and Q intersect at right angles, then locus of (h, k) will be director circle of x2 + y2 = 2a2 i.e. x2 + y2 = 4a2. PQ is chord of contact of the circle x2 + y2 = 2a2 w.r.t. the point (h, k) i.e. equation of PQ is hx + ky = 2a2 ……(2) (1) and (2) are same equation cos2θ + sin2 θ = 1 ⇒ h2 + k2 = 4a2 ∴ locus of (h,k) is x2 + y2 = 4a2 which is director circle to x2 + y2 = 2a2 Illustration 11: If tangent at (1, 2) to the circle x2+ y2 = 5 intersects the circle x2+ y2 = 9 at A and B and tangents at A and B to the second circle meet at point C, then find coordinates of C. Solution: The tangent at (1, 2) is given by x + 2y – 5 = 0 Let point C be (h, k). Then this tangent will be chord of contact of this point w.r.t. IInd circle so, equation will be xh + yk – 9 = 0, both are same lines so Illustration 12: Tangents PA and PB are drawn to x2+y2=4 from the point P(3,0). Find the area of triangle PAB. Solution: Equation of AB is T = 0 i.e. 2.2 FAMILY OF CIRCLES (i) If S ≡ x2 + y2 + 2gx + 2fy + c = 0 and S′ ≡ x2 + y2 + 2g′x + 2f′y + c′ = 0 are two intersecting circles, then family of circles passing through the point of intersection of S and S’ is given by S + λS′ = 0, (where λ is a parameter λ ≠ –1). (ii) If S ≡ x2 + y2 + 2gx + 2fy + c = 0 is a circle which is intersected by the straight line L=lx+my+n=0 at two real and distinct points, then the equation of the family of circles passing through the point of intersection of circle and the given line is given by S + λL = 0.(whereλ is a parameter). (iii) The equation of a family of circles passing through two given points (x1, y1) and (x2, y2) can be written in the form. (x – x1)(x – x2) + (y – y1)(y – y2) + (where λ is a parameter). (iv) The equation of the family of circles which touch the line y – y1 = m(x – x1) at (x1, y1) for any value of m is (x – x1)2 + (y – y1)2 + λ[(y – y1) –m(x – x1)] = 0. If m is infinite, the equation is (x – x1)2 + (y – y1)2 + λ(x – x1) = 0. Illustration 13: The equation of the common chord of two circles is x+ y = 1. One of the circles has the ends of a diameter at the points (1 , -3) and (4, 1) and the other passes through the point (1 ,2). Find the equation of the two circles Solution: Let c1 is the circle whose end point of a diameter are given ∴ c1 ≡ (x – 1) (x – 4) + (y +3) ( y – 1) = 0 ≡ x2 + y2 – 5x + 2y + 1 = 0 Let c2 is another circles ∴ Equation of family of circles passing through the points of intersection of circle c1 and the given line x2 + y2 – 5x + 2y +1 + λ(x + y – 1) = 0 …………….(1) for c2 , 1st equation satisfy the point (1 ,2 ) ∴ λ = ∴ c2 = 2x2 + 2y2 – 15x – y + 7 = 0 2.3 RADICAL AXIS The radical axis of two circles is the locus of a point from which the tangent segments to the two circles are of equal length. Equation to the Radical Axis Consider S ≡ x2 + y2 + 2gx + 2fy + c = 0 and S' ≡ x2 + y2 + 2g'x + 2f'y + c' =0, then S-S’=0 represents the equation of the Radical Axis to the two circles i.e. 2x(g – g′) + 2y(f – f′) + c – c′ = 0. Note: 1. If S = 0 and S' = 0 intersect in real and distinct points then S – S' = 0 is the equation of the common chord of the two circles. ⋅ 2. If S' = 0 and S = 0 touch each other, then S – S' = 0 is the equation of the common tangent to the two circles at the point of contact. 3. To find the equation of the radical axis of two circles, first make the coefficients of x2 and y2 in the equation of the two circles equal to unity. 4. The radical axis of two circles is perpendicular to the line joining their centres. 5. The radical axis of three circles taken in pairs meet at a point , called the radical centre of the circles. Coordinates of radical centre can be found by solving the equations S1=S2=S3=0. 6. The radical centre of three circles described on the sides of a triangle as diameters is the orthocentre of the triangle. 7. If two circles cut a third circle orthogonally, then the radical axis of the two circles pass through the centre of the third circle. or the locus of the centre of a circle cutting two given circles orthogonally is the radical axis of the two circles. 8. The radical axis of the two circles will bisect their common tangents. Illustration 14: If the circle x2+y2+2a1x+2b1y+c1=0 bisects the circumference of x2+y2+2a2x+2b2y+c2=0, then show that 2a2 (a1 – a2)x + 2b2(b1 – b2)y + c1 – c2 = 0 Solution: Clearly the centre of second circle i.e., (–a2, –b2) should lie on the common chord of a circles i.e., on the line 2 (a1 – a2)x + 2(b1 – b2)y + c1 – c2 = 0 Hence 2a2 (a1 – a2)x + 2b2(b1 – b2)y + c1 – c2 = 0 3. SOLVED OBJECTIVE PROBLEMS 1: Equation of chord AB of circle x2 + y2 = 2 passing through P(2 , 2) such that PB/PA = 3, is given by (a) x = 3 y (b) x = y (c) y – 2 = (x – 2) (d) none of these Solution: Key concept: Using the concept of parametric equation of any a line. Any line passing through (2, 2) will be of the form = r If this line cuts the circle x2+y2=2 , then (rcosθ+2)2 +(r sinθ+2)2 =2 ⇒ r2 + 4(sinθ+ cosθ)r +6 = 0. This is a quadratic equation in r, roots of this equation will represent distance PB and PA. Let the roots be r1 and r2. Then , now if r1 = α, r2 = 3α, then 4α = - 4(sinθ + cosθ), 3α2 = 6 ⇒ sin2θ = 1⇒ θ = π/4 . So required chord will be y – 2 = 1 ( x –2) ⇒ y = x. Alternative solution: Key concept: Using the basic property of a circle. PA.PB = PT2 = 22 + 22 – 2 = 6 . . . . (1) . . . . (2) From (1) and (2), we have PA = , PB =3 ⇒ AB = 2 . Now diameter of the circle is 2 (as radius is ) Hence line passes through the centre ⇒ y = x . Hence (B) is the correct answer. 2: If the line y – 2 = m (x – 1) cuts the circle x2 + y2 = 9 at two real point, then the number of possible values of m is (a) 1 (b) 2 (c) infinite (d) 0 Solution: Since the given line always passes through (1, 2), the interior point of the circle, so m can have any real value. Hence (C) is the correct answer. 3: The range of values of  for which the circle x2 + y2 = 4 and x2 + y2 – 4x + 9 = 0 have two common tangent is (a) (b) (c) (d) none Key concept: There will exist two common tangents when both the circles are intersecting. Hence apply the condition. Solution: Solving the equation 4 – 4x + 9 = 0 ⇒ . It should have two real and distinct values so Hence (A) is the correct answer. 4: The equation of a circle of radius 1 touching the circles x2 + y2 – 2⏐x⏐ = 0 is (a) (b) (c) (d) Approach: First simplify the equation of the given circle and then plot its graph. Solution: The given circles are x2 +y2 – 2x = 0 , x > 0 and x2 +y2 + 2x = 0, x < 0 From the figure the centre of the required circle will be . Hence (B) is the correct answer. 5: Radius of smaller circle that touches the line y  x ot (1, 1) and also touches the x–axis is : (a) 2 – (b) 2 + (c) – 1 (d) 1 + Solution: Since OA  OP ⇒ p ∈ ( , 0) equation of AC (y – 1)  –1 (x – 1) i.e. y  –x + 2 equation of PC x  ⇒ C  ( , 2 – ) Thus radius is 2 – Hence (A) is the correct answer. 6: A circle touches the lines and has unit radius. If the centre of this circle lies in the first quadrant then one possible equation of this circle is (a) (b) (c) (d) Solution: Angle between lines is 60° – 30° = 30°. Thus equation of their acute angle bisector is y tan (30° + 15°). x i.e., y = x . let C = (h, h) then thus equation of circle is Hence (C) is the correct answer. 7: If the circles and have exactly two common tangents, then (a) 1<|a|<8 (b) 2<|a|<8 (c) 3<|a|<8 (d) 4<|a|<8 Solution: Centres of the circles are (4, –1) and (1, 3). Their radii are 3 and ⏐a⏐ respectively. They will have exactly two common tangents if they meet in two distinct points. That means From we get a ∈ From we get a ∈ (–8, 8) Hence (B) is the correct answer. 8: If (a, 0) is a point on a diameter of the circle x2+y2 =4, then x2 – 4x – a2 =0 has (a) exactly one real root in ( -2, -1] (b) exactly one real root in [2, 5] (c) distinct roots greater than 1 (d) distinct roots less than -1 Solution: Since (a, 0) is a point on the diameter of the circle x2 +y2 = 4, so maximum value of a2 is 4 Let f(x) = x2 – 4x – a2 clearly f(-1) = 5 – a2 > 0, f(2) = -(a2 + 4) < 0 f(0) = -a2 < 0 and f( 5) = 5 - a2 > 0 so graph of f(x) will be as shown Hence (A) is the correct answer. 9. If A is a point on the circle x2 + y2 – 4x +6y –3 = 0 . which is farthest from the point ( 7,2) , then (a) (2 - 2 (b) (2 + 2 (c) (2 + 2 (d) (2 - 2 Solution: Given circle is (x – 2)2 + (y + 3)2 = 16 slope of PA = Parametric equation of line PC is Hence the point is Hence (A) is the correct answer. 10. Length of the common chord of circles x2 + y2 –2x –4y +1 =0 and x2 + y2 –4x –2y +1 =0 is (a) (b) (c) 4 (d) Solution: Equation of common chord is s1 – s2 = 0 2x – 2y = 0 ⇒ x – y = 0 Length of chord = Hence (A) is the correct answer. 11. C1 be a circle of radius one unit, touching both axes. C2 is another circle that also touches both axes and also touches C1 externally, then radius of C2 is, (a) 2 (b) 3 + 2 (c) 3 - (d) 3 + Solution: Equation of C1 (x – 1)2 + (y – 1)2 = 1 Let equation of C2 (x – α)2 + (y – α)2 = α2 Both the circle touches each other Hence (B) is the correct answer. 12. The equation of the circle of radius which touches the line x + y = 1 at (2, –1) is (a) x2 + y2 – 4x +2y+ 3= 0 (b) x2 + y2 + 6x +7= 0 (c) x2 + y2 – 2x +4y+ 3= 0 (d) none of these Solution: Family of circle touches x + y – 1 = 0 at (2, –1) is (x – 2)2 + (y + 1)2 + λ(x + y – 1) = 0 Now radius of this circle is Hence Hence (C) is the correct answer. 13. If a circle passes through the points of intersection of the co-ordinate axes with the lines λ x – y + 1 = 0 and x – 2y + 3 = 0, then the value of λ is (a) 2 (b) 4 (c) 6 (d) 3 Solution: If and x – 2y + 3 = 0 intersect the axes at con cyclic points then λ.1 = (–1)(–2) λ =2 Hence (A) is the correct answer. 14. A square is inscribed in the circle x2 + y2 –2x + 4y +3 = 0. Its sides are parallel to the coordinate axes. Then one vertex of the square is (a) (b) (c) (d) none of these Solution: Two diagonals will be inclined at 45° with the axes. Parametric equation Point (2, –1) (0, –3) Parametric equation other diagonal Hence points are (0, –1), (2, –3) Hence (D) is the correct answer. 15. If (a, 0) is a point on a diameter of the circle x2+y2 =4, then x2 – 4x – a2 =0 has (a) exactly one real root in ( -2, -1] (b) exactly one real root in [2, 5] (c) distinct roots greater than 1 (d) distinct roots less than -1 Solution: Since (a, 0) is a point on the diameter of the circle x2 + y2 = 4 So maximum value of a2 is 4. Let f(x) = x2 – 4x – a2 clearly f(–1) = 5 – a2 > 0 f(5) f(0) =– a2 > 0 and f(5) = 5 – a2 > 0 So graph of f(x) will be as shown Hence (A), (B), (C) and (D) are the correct Hence (A, B, C, D) is the correct answer. 16. A line 2x + 3y = 5 intersect the axes at A and B. Locus of the centre of the family of the circle passing through A and B is (a) 36x + 24y + 25 = 0 (b) 24x + 36y + 25 = 0 (c) 36x – 24y – 25 = 0 (d) 24x – 36y – 25 = 0 Solution: Locus of the centre of the family of the circle passing through A and B, will be perpendicular side bisector of AB Hence (C) is the correct answer. 17. Equation of circle touching the lines will be (a) (x –2)2 +(y – 3)2 = 12 (b) (x –2)2 +(y – 3)2 = 4 (c) (x –2)2 +(y – 3)2 = 4 (d) (x –2)2 +(y – 3)2 = 8 Solution: Clearly |x – 2| + |y – 3| =4 will be a square. Hence the centre of circle will be (2, 3) and radius . Hence equation of circle is (x – 2)2 + (y – 3)2 = 8 Hence (D) is the correct answer. 18. Equation of a circle that cuts the circle x2 + y2 + 2gx + 2fy + c = 0, lines x = -g and y = -f orthogonally, is; (a) x2 + y2 + 2gx + 2fy + g2 + f2 –c = 0 (b) x2+ y2 + 2gx + 2fy + g2 + f2 + c= 0 (c) x2 + y2 + 2gx + 2fy - g2 - f2 –c = 0 (d) none of these Solution: Lines x = – g and y = – f will be diameter for the circle. Equation of circle will be (x + g)2 + (y + f)2 = r2 . Now this intersect x2 + y2 + 2gx + 2fy + c = 0 orthogonally Hence 2g.g. +2f.f = c + k ⇒ k = 2g2 + 2f2 – c Hence (D) is the correct answer. 19. If a chord of circle x2 + y2 –4x –2y – c =0 is trisected at the points (1/3 ,1/3) and (8/3 ,8/3) , then (a) c = 10 (b) c = 15 (c) c = 20 (d) c = 5 Solution: Clearly mid point of chord is and length of chord will be . In c = 20 Hence (C) is the correct answer. 20. The circle for which the line joining the points (am2, 2am) and is a diameter, is touched, for all values of m, by the line (a) x= a (b) x+a =0 (c) x= 2a (d) x+2a= 0 Solution: Both the points (am2, 2am) and are lying on the parabola y2 = 4ax, and are end points of any focal chord. Hence circle which touch at directrix which is x = –a Hence (A) is the correct answer. 21. A circle passing through the points touches the pair of lines x2-y2-2x +1 =0. The centre of the circle is (a) (4, 0) (b) (6, 0) (c) (0, 4) (d) (5, 0) Solution: Given pair of lines are (x – 1)2 = y2 ⇒ x – 1 = y and x + y –1 = 0 The centre will lie on x–axis (α, 0) Hence (A, B) is the correct answer. 22. The locus of the centre of a circle which touches externally the circle x2 + y2 – 6x – 6y – 14 = 0 and also touches the y-axis is given by the equation: (a) x2 – 6x – 10y + 14 = 0 (b) x2 – 10x – 6y + 14 = 0 (c) y2 – 6x – 10y + 14 = 0 (d) y2 – 10x – 6y + 14 = 0 Solution: Equation of circle touches y–axis is given by (x – 4)2 + (y – k)2 = h2 Given by (x – 4)2 + (y – k)2 = h2 this touches (x – 3)2 + (y – 3)2 = 4 Hence locus is y2 – 6y – 10x + 14 = 0 Hence (D) is the correct answer. 23. The equation of circle which touches the line 5x + 12 y =1 and which has its centre at (3, 4) is: (a) (x – 3)2 + (y – 4)2 = (b) (x – 3)2 + (y – 4)2 = (c) (x – 3)2 + (y – 4)2 = (d) none of these Solution: Since the circle whose centre (3, 4) touches the line 5x + 12y =1, radius of this circle will be Hence equation is (x – 3)2 + (y – 4)2 = Hence (C) is the correct answer. 24. The intercept on the line y = x by the circle x2 + y2 – 2x = 0 is AB. Equation of the circle with AB as a diameter is: (a) x2 + y2 + x + y = 0 (b) x2 + y2 – x – y = 0 (c) x2 + y2 + x – y = 0 (d) none of these Solution: The line x = y intersect the circle x2 + y2 – 2x = 0 at (0, 0) and (1, 1) Hence equation of required circle is (x – 0)(x –1) + (y – 0)(y – 1) = 0 Hence (B) is the correct answer. 25. The number of common tangents to the circles x2 + y2 = 4 and x2 + y2 – 6x – 8y = 24 is: (a) 0 (b) 1 (c) 3 (d) 4 Solution: Both the circles are touching internally only one common tangent is possible Hence (B) is the correct answer. 26. Let L1 be a straight line passing through the origin and L2 be the straight line X + Y = 1. If the intercepts made by the circle x2 + y2 – x + 3y = 0 on L1 and L2 are equal, then which of the following equations can represent L1: (a) x + y = 0 (b) x – y = 0 (c) x + 7y = 0 (d) x – 7y = 0 Solution: Given PP′ = QQ′ ⇒ (PP′)2 = (QQ′)2 ⇒ {(OP′)2 – (op)2} = (OQ′)2 – (OQ)2 ⇒ OP = OQ ⇒ ⇒ m = 1, –1/7 ⇒ lies are y = x, y = – Hence (B, C) is the correct answer. 27. The equation of the radical axis of the two circles 7x2 + 7y2 – 7x + 14y + 18 = 0 and 4x2 + 4y2 – 7x + 8y + 20 = 0 is given by (A) 3x2 + 3y2 – 6y – 2 = 0 (B) 21x – 68 = 0 (C) x – 2y – 5 = 0 (D) None of these. Solution: Here, and Now, This shows that these circles touch each other internally. Hence (B) is the correct answer. 27. If the circles x2 + y2 + 2x + 6ky + 6 = 0 and x2 + y2 + 2ky + k = 0 intercept orthogonally, then k is: (a) (b) (c) (d) Solution: Sine both the circle are intersecting orthogonally hence apply the condition 2g1g2 + 2f1f2 = c1 + c2 Hence (A) is the correct answer. 28. If the tangent at a point P on the circle x2 + y2 + 6x + 6y = 2 meets the straight line 5x – 2y + 6 = 0 at a point Q on y-axis, then the length PQ is: (a) 4 (b) (c) 5 (d) Solution: x2 + y2 + 6x + 6y – 2 = 0 The line 5x – 2y + 6 = 0 meets the y–axis at Q(0, 3) Length of tangent drawn from Q, i.e. Hence (C) is the correct answer. 29: Tangents to the circle x2 + y2 = a2 cut the circle x2 + y2 = 2a2 at P and Q. The tangents at P and Q to the circle x2 + y2 = 2a2 intersect at (A) right angles (B) 60° (C) can't be determined (D) none of the above. Solution: Equation of tangent at any point (a cosθ, a sin θ) is x cosθ + ysin θ = a ……(1) Let the point of intersection of the tangents at P and Q be (h, k). If the tangents at P and Q intersect at right angles, then locus of (h, k) will be director circle of x2 + y2 = 2a2 i.e. x2 + y2 = 4a2. PQ is chord of contact of the circle x2 + y2 = 2a2 w.r.t. the point (h, k) i.e. equation of PQ is hx + ky = 2a2 ……(2) (1) and (2) are same equation cos2θ + sin2 θ = 1 ⇒ h2 + k2 = 4a2 ∴ locus of (h,k) is x2 + y2 = 4a2 which is director circle to x2 + y2 = 2a2 Hence (A) is the correct answer. 30: If touches the circle x2+ y2 = r2, then prove that the point lies. (A) on a circle (B) in a circle (C) on a straight line (D) none of these. Solution: touches x2+y2 =r2 ⇒ ⇒ ⇒ lies on x2+y2 = , which is a circle. Hence (A) is the correct answer. 31: The locus of the mid–points of the chords of the circle x2 + y2 – 2x – 6y – 10 = 0 which pass through the origin is (A) x2 + y2 + x – 3y = 0 (B) x2 + y2 – x – 3y = 0 (C) x2 + y2 – x + 3y = 6 (D) x2 + y2 – x – 3y = 6 Solution: Let (h, k) be the coordinates of the mid point. Equation of the chord whose mid point is (h, k) is xh + yk – (x + h) – 3(y + k) – 10 = h2 + k2 – 2h – 6k – 10 (using T = S1) i.e., h2 + k2 – h(x + 2) – k(6 + y) + x + h + 3y + 3k = 0 This chord passes through the origin (0, 0). Hence h2 + k2 – h – 3k = 0 Hence the required locus is x2 + y2 – x – 3y = 0. Alternative: If m1 is the slope of the chord, then If m2 is the slope of the perpendicular to the chord then where (1, 3) is the centre of the given circle. But m1m2 = –1 ⇒ h2 + k2 – h – 3k = 0 Hence the required locus is x2 + y2 – x – 3y = 0. Hence (B) is the correct answer. 32: The equation of the circle with centre on the line 2x + y = 0 and touching the lines 4x – 3y + 10 = 0 and 4x – 3y – 30 = 0 is (A) (x – 1)2 + (y + 2)2 = 16 (B) (x – 2)2 + (y + 4)2 = 16 (C) (x + 1)2 + (y – 2)2 = 16 (D) none of these. Solution: Since the two lines are tangents to the given circle and they are parallel, we have Radius = = 4 units (we know that if ax + by + c = 0 and ax + by + c′ = 0 are two parallel tangents to a circle, then equation of the line parallel to given lines and passing through the centre is given by ax + by + ) Hence the centre of the circle lies on 4x – 3y – 10 = 0 Also the centre lies on 2x + y = 0 Hence the coordinates of the centre are (1, –2). ⇒ The equation of the circle is (x – 1)2 + (y + 2)2 = 16. Hence (A) is the correct answer. 33: One of the diameters of the circle circumscribing the rectangle ABCD is 4y = x + 7. If A and B are (–3, 4), (5, 4), the area of the rectangle is (A) 16 sq. units (B) 24 sq. units (C) 32 sq. units (D) 48 sq. units Solution: Let O(h, k) be the centre of the circle ⇒ (h + 3)2 = (h – 5)2 ⇒ h2 + 6h + 9 = h2 – 10h + 25 ⇒ 16h = 16 ⇒ h = 1 (1, k) lies on 4y = x + 7 ⇒ k = 2 OB = Let E be the mid–point of AB Then E is (1, 4) ⇒ OE = 2 ⇒ AB = 8 and BC = 4 ⇒ Area = 4 × 8 = 32 sq. units. Hence (C) is the correct answer. 34: Four distinct points (2K, 3K), (1,0), (0,1) and (0,0) lie on a circle when (A) all are integral values of K (B) 0 < K < 1 (C) K < 0 (D) For two values of K Solution: The equation of the circle passing through the point (1, 0), (0, 1) and (0, 0) is x2 + y2 – x – y = 0, which passes through (2k, 3k), if 4k2 + 9k2 – 2k – 3k = 0 ⇒ k = 0, k = . Hence (D) is the correct answer. 35: The equation of the image of the circle x2 + y2 + 16x – 24 y + 183 = 0 by the line mirror 4x + 7y + 13 = 0 is (A) x2 + y2 + 32x – 4y + 235 = 0 (B) x2 + y2 + 32x + 4y – 235 = 0 (C) x2 + y2 + 32x – 4y –235 = 0 (D) x2 + y2 + 32x + 4y + 235 = 0 Solution: Image of the centre = (–16, –2) So equation of required circle is (x + 16)2 + (y +2)2 = 52 Hence (D) is the correct answer. 36: If the circles x2 + y2 + 2x + 2ky +6 = 0 and x2 + y2 + 2ky + k = 0 intersect orthogonally, then k is (A) 2 or – (B) –2 or – (C) 2 or (D) –2 or Solution: Apply 2gg1 + 2ff1 = c1 + c2 ⇒ 2(0) + 2kk = 6+ k or 2k2 – k – 6 = 0 ⇒ k = 2, –3/2 Hence (A) is the correct answer. 37: The equation x2 + y2 + 4x + 6y + 13 = 0 represents (A) a circle (B) a pair of straight lines (C) a point (D) none of these Solution: (x + 2)2 + (y + 3)2 = 0 ⇒ x + 2 = 0, y + 3 = 0. Hence the point (–1, –2) Hence (C) is the correct answer. 38: Equation of the circle whose radius is 5 and which touches externally the circle x2 + y2 – 2x – 4y – 20 = 0 at the point (5, 5) is (A) (x – 9)2 + (y – 6)2 = 52 (B) (x – 9)2 + (y – 8)2 = 52 (C) (x – 7)2 + (y – 3)2 = 52 (D) none of these Solution: If (h, k) be the centre then (5, 5) is the mid–point of (h, k) and (1, 2) because the radii of both the circles are 5 each ∴ h = 9, k = 8. Hence (B) is the correct answer. 39: The angle between tangents from the origin to the circle (x –7)2 + (y + 1)2 = 25 is (A) π/3 (C) π/6 (C) π/2 (D) 0 Solution: sinθ = ⇒ θ = Angle between tangents = . Hence (C) is the correct answer. 40: If the circles (x –a)2 + (y –b)2 = c2 and (x –b)2 + (y –a)2 = c2 touch each other then (A) a = b ± 2c (B) a = b ± c (C) a = b ± c (D) none of these Solution: The circles will touch each other ∴ c1c2 = r1 + r2 ⇒ a = b ± 2 Hence (B) is the correct answer. 41: A circle of radius a with both coordinates of its centre positive, touches the axis of x and the straight line 3y = 4x. then its equation is (A) x2 + y2 –4ax –2y –4a2 = 0 (B) x2 + y2 –4ax –2ay + 4a2 = 0 (C) x2 + y2 + 4ax + 2ay –4a2 = 0 (D) none of these Solution: Since it touches x–axis, ordinate of its centre = radius ⇒ (h, a) where both h and a positive Now apply condition of tangency with 3y = 4x. Hence (B) is the correct answer. 42: The shortest distance between the circles x2 +y2 =1 and x2 +y2 –10x –10y+ 41=0 is (A) –1 (B) 0 (C) (D) 5 Solution: The distance between the centres of the circles is 5 . Sum of the radii = 1 + 3 = 4. Hence, shortest distance = 5 –4. Hence (D) is the correct answer. 43: The length of the common chord of the circles x2 + y2 + 2x + 3y + 1 = 0 and x2 + y2 + 4x + 3y + 2 = 0, is (A) (B) 2 (C) 3 (D) Solution: Equation of common chord is 2x + 1 = 0 ⇒ x = –1/2 This cuts the circle at points + y2 + 2 + 3y + 1 = 0 ⇒ + y2 – 1 + 3y + 1= 0 ⇒ 4y2 + 12y + 1 = 0 ⇒ y = ∴ length = 2 . Hence (B) is the correct answer. 44: The equation of the circle touching the line 2x + 3y + 1 = 0 at the point (1, –1) and passing through the focus of the parabola y2= 4x is (A) 3x2 + 3y2 – 8x + 3y + 5 = 0 (B) 3x2 + 3y2 + 8x – 3y + 5 = 0 (C) x2 + y2 – 3x + y + 6 = 0 (D) none of these Solution: Equation of the circle is λ (2x + 3y + 1) + (x – 1)2 + (y + 1)2 = 0 It passes through (1, 0) ⇒ λ = Equation of circle is 3x2 + 3y2 – 8x + 3y + 5 = 0 Hence (A) is the correct answer. 45: The circles x2 + y2 + 6x + 6y = 0 and x2 + y2 –12x –12y = 0 (A) touch each other internally (B) touch each other externally (C) intersect in two points (D) cut orthogonally Solution: Centres of circles are (–3, –3) and (6, 6) and radii are 3 and 6 . So C1C2 = r1 + r2 Hence (A) is the correct answer. 46: Equation of a circle with centre (4, 3) touching the circle x2+ y2 = 1 is (A) x2+y2–8x–6y–9 = 0 (B) x2+y2–8x–6y+11 = 0 (C) x2+y2–8x–6y–11 = 0 (D) x2+y2–8x–6y+9 = 0 Solution: Let the circle touching the circle x2 + y2 = 1, be x2 + y2 – 8x – 6y + k = 0, The equation of the common tangent is S1 – S2 = 0 i.e. 8x + 6y–1–k = 0 This is a tangent to the circle x2+y2 = 1. Hence ± 1 = ⇒ k+1 = ± 10 ⇒ k = –11 or 9 Therefore the circles are x2+y2–8x–6y+9 = 0 and x2+y2–8x–6y–11 = 0 Hence (C) and (D) are the correct answers. 47: The number of common tangents that can be drawn to the circle x2+ y2– 4x – 6y – 3 = 0 and x2 + y2 + 2x + 2y + 1 = 0 is (A) 1 (B) 2 (C) 3 (D) 4 Solution: The two circles are x2 + y2 – 4x – 6y – 3 = 0 and x2 + y2 + 2x + 2y + 1 = 0 Centre: C1 ≡ (2, 3), C2 ≡ (–1, –1) radii: r1 = 4, r2 = 1 We have, C1 C2 = 5 = r1 + r2, therefore there are 3 common tangents to the given circles. Hence (C) is the correct answer. 48: The tangents drawn from the origin to the circle x2+y2–2rx–2hy+h2 = 0 are perpendicular if (A) h = r (B) h = –r (C) r2+h2 = 1 (D) r2=h2 Solution: The combined equation of the tangents drawn from (0,0) to x2+y2–2rx–2 hy + h2 = 0 is (x2+y2 – 2 rx–2 hy + h2)h2 = (–rx – hy + h2)2 This equation represents a pair of perpendicular straight lines If coefficient of x2 + coefficient of y2 = 0 i.e. 2h2 – r2 – h2 = 0 ⇒ r2 = h2 or r = ± h. Hence (A), (B), and (D) are the correct answers. 49. Area of the circle in which a chord of length makes an angle π/2 at the centre is (A) π/2 (B) 2π (C) π (D) π/4 Solution: Let AB be the chord of length , O be the centre of the circle and let OC be the perpendicular from O on AB. Then AC = BC = In ΔOBC, OB = BC cosec 45° =  Area of the circle = π (OB)2 = π. Hence (C) is the correct answer. 550. If the lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 cut the coordinate axes in concyclic points, then (A) a1a2 = b1b2 (B) a1b1 = a2b2 (C) a1b2 = a2b1 (D) None of these. Solution: Let the given lines be L1 ≡ a1x + b1y +c1 = 0 and L2 ≡ a2x + b2y + c2 = 0. Suppose L1 meets the coordinate axes at P and Q and L2 meets at R and S. Then the coordinates of P, Q, R, S are Since P, Q, R, S are concyclic, therefore, OP . OR = OQ . OS . Hence (A) is the correct answer. 751. The lines 2x – 3y + 5 = 0 and 3x – 4y = 7 are diameters of a circle of area 154 sq. units, then the equation of the circle is (A) x2 + y2 + 2x – 2y – 62 = 0 (B) x2 + y2 + 2x – 2y – 47 = 0 (C) x2 + y2 – 2x + 2y – 47 = 0 (D) x2 + y2 – 2x + 2y – 62 = 0 Solution: The centre of the required circle lies at the intersection of 2x – 3y – 5 = 0 and 3x – 4y – 7 = 0. Thus, the coordinates of the centre are (1, –1) let r be the radius of the circle. Then, by hypothesis πr2 = 154 ⇒ r = 7 Hence, the equation of the required circle is Hence (C) is the correct answer. 852. The tangents to x2 + y2 = a2 having inclinations α and β intersect at P. If cot α + cot β =0, then the locus of P is (A) x + y = 0 (B) x – y = 0 (C) xy = 0 (D) None of these. Solution: Let the coordinates of P be (h, k). Let the equation of a tangent from P(h, k) to the circle x2 + y2 = a2 be Since, P(h, k) lies on , therefore, The is a quadratic in m. Let the two roots be m1 and m2, then But tan α = m1, tan β = m2 and it s given that cot α + cot β = 0. ⇒ hk = 0. Hence, the locus of (h, k) is xy = 0. Hence (C) is the correct answer. 1553. The area of circle centred at (1, 2) and passing through (4, 6) is (A) 5 π (B) 10 π (C) 25 π (D) None of these. Solution: Centre (1, 2) passes through (4, 6) radius =  Area = πr2 = 25 π. Hence (C) is the correct answer. 1854. The pole of the straight line 9x + y – 28 = 0 with respect to the circle 2x2 + 2y2 – 3x + 5y – 7 = 0 is (A) (3, 1) (B) (1, 3) (C) (3, – 1) (D) (–3, 1) Solution: Let (h, k) be the pole of 2x2 + 2y2 – 3x + 5y – 7 = 0 ⇒ The polar is 4hx + 4ky – 3(x + h) + 5(y + k) – 14 = 0 ⇒ (4h – 3)x + (4k + 5)y – (3h – 5k + 14) = 0 Comparing with 9x + y – 28 = 0 4h – 3 = 9 ⇒ h = 3 and 4k + 5 = 1 ⇒ k = – 1 and 3h – 5k + 14 = 28 put h = 3 and k = – 1 9 + 5 + 14 = 28 = R.H.S. ⇒ (3, – 1) is the pole. Hence (C) is the correct answer. 2555. If the line y = x + 3 meets the circle x2 + y2 = a2 at A and B, then equation of the circle on AB as diameter is (A) x2 + y2 + 3x – 3y – a2 + 9 = 0 (B) x2 + y2 – 3x + 3y – a2 + 9 = 0 (C) x2 + y2 + 3x + 3y – a2 + 9 = 0 (D) None of these. Solution: Use S + λP = 0. Its centre lies on y = x + 3  λ = 3. Hence (A) is the correct answer. 3056. The equation of the circumcircle of the triangle formed by the lines and y = 0 is (A) x2 + y2 – 4y = 0 (B) x2 + y2 + 4x = 0 (C) x2 + y2 – 4y = 12 (D) x2 + y2 + 4x = 12 Solution: The three vertices by solving in pairs are (0, 6), ( , 0) and (– , 0) if the circle be g, f, c, then g = 0, f = – 2, c = – 12. Hence (C) is the correct answer. 3757. Polar of origin (0, 0) w.r.t. the circle x2 + y2 + 2λx + 2μy + c = 0 touches the circle x2 + y2 = r2, if (A) c = r (λ2 + μ2) (B) r = c (λ2 + μ2) (C) c2 = r2 (λ2 + μ2) (D) r2 = c2 (λ2 + μ2) Solution: Polar of (0, 0) is x . 0 + 0 . y + λ (x + 0) + μ(y + 0) + c = 0 λx + μy + c = 0 . . . (1) (1) will touch the circle if the distance of origin from (1) = r . Hence (B) is the correct answer. 1158. Image of the circle x2 + y2 – 6x + 8 = 0 in the line y = x is (A) x2 + y2 + 6y + 8 = 0 (B) x2 + y2 – 6y – 8 = 0 (C) x2 + y2 – 6y + 8 = 0 (D) None of these. Solution: Centre and radius of the circle is (3, 0) and 1. Image of centre (3, 0) in the line x – y = 0 is (0, 3). Therefore, image of the given circle in the line y = x is . Hence (C) is the correct answer. 2159. Radius of the smallest circle passing through the point of intersection of circles x2 + y2 + 2x – 3 and x2 + y2 + 3x – y = 0 is (A) 1 (B) (C) (D) None of these. Solution: . . . (1) . . . (2) Equation of the common chord is x – y + 3 = 0 . . . (3) Perpendicular distance of P(–1, 0) centre of circle (1), from the common chord (3) is and radius of the circle (1) is  Radius of the smallest circle is . Hence (B) is the correct answer. 2560. Locus of the poles of the circle x2 + y2 = a2 so that polar is always touching the circle x2 + y2 = b2 is (A) a2 (x2 + y2) = b4 (B) b2 (x2 + y2) = a4 (C) x2 + y2 = a2 + b2 (D) None of these. Solution: Let pole be (h, k), then equation of the polar is . . . (1) Since (1) is touching the circle , therefore  Required locus is . Hence (AB) is the correct answer. 3461. AB is a diameter of a circle and C is any point on the circumference of the circle. Then (A) the area of ΔABC is maximum when it is isosceles (B) the area of ΔABC is minimum when it is isosceles (C) the perimeter of ΔABC is maximum when it is isosceles (D) None of these. Solution: ΔABC will always be right angled at C. Let AC = x, BC = y and area For maxima/minima, Similarly,  Triangle is isosceles hence area is maximum. Hence (A) is the correct answer. 6562. If the straight line y = mx is outside the circle x2 + y2 – 20y + 90 = 0, then (A) m < 3 (B) | m | < 3 (C) m > 3 (D) | m | > 3 Solution: Given equation can be written as or which is impossible, as left hand term is always positive term. Hence (AD) is the correct answer. 80. The equation of the radical axis of the two circles 7x2 + 7y2 – 7x + 14y + 18 = 0 and 4x2 + 4y2 – 7x + 8y + 20 = 0 is given by (A) 3x2 + 3y2 – 6y – 2 = 0 (B) 21x – 68 = 0 (C) x – 2y – 5 = 0 (D) None of these. Solution: Hence (A) is the correct answer.