CIRCLE SYSTEM-02- (THEORY)-PART-II

Condition Position Diagram No. of common tangents (i) Do not intersect or one outside the other 4 (ii) One inside the other 0 (iii) External touch 3 (iv) Internal touch 1 (v) Intersection at two real points 2 Example : 32 If circles and touch each other, then (a) (b) (c) (d) Solution : (d) ; ; Circles touch each other, therefore  Multiplying by we get . Example : 33 If two circles and intersect in two distinct points, then (a) (b) (c) (d) Solution : (a) When two circles intersect each other, then difference between their radii < Distance between their centres …..(i) Sum of their radii > Distance between their centres …..(ii) Hence by (i) and (ii), 2 < r < 8. Example : 34 The equation of the circle having the lines as its normals and having size just sufficient to contain the circle is (a) (b) (c) (d) Solution : (b) Given pair of normals is or Normals are and The point of intersection of normals and is the centre of required circle, we get centre and other circle is or …..(i) Its centre and radius Since the required circle just contains the given circle (i), the given circle should touch the required circle internally from inside. Therefore, radius of the required circle Hence, equation of required circle is or . Example : 35 The equation of the circle which touches the circle externally and to which the lines are normals, is (a) (b) (c) (d) Solution : (d) Joint equations of normals are Given normals are and , which intersect at centre of circle whose coordinates are (3, 1). The given circle is ; If the two circles touch externally, then Equation of required circle is Example : 36 The number of common tangents to the circles and is (a) 0 (b) 1 (c) 3 (d) 4 Solution : (b) Circles and Centres and radii , i.e. circles touch internally. Hence there is only one common tangent. Example : 37 There are two circles whose equations are and If the two circles have exactly two common tangents, then the number of possible values of n is (a) 2 (b) 8 (c) 9 (d) None of these Solution : (c) For , the centre = (0, 0) and the radius = 3 For . The centre = (4, 3) and the radius or or Circles should cut to have exactly two common tangents. So, , or or or Therefore, common values of n should satisfy But , So, . Number of possible values of n = 9. 4.18 Common chord of two Circles. (1) Definition : The chord joining the points of intersection of two given circles is called their common chord. (2) Equation of common chord : The equation of the common chord of two circles ….(i) and ….(ii) is i.e. (3) Length of the common chord : Where radius of the circle and length of the perpendicular from the centre to the common chord PQ. Note :  The length of the common chord is where p1 and p2 are the lengths of perpendicular drawn from the centre to the chord.  While using the above equation of common chord the coefficient of and in both equation should be equal.  Two circle touches each other if the length of their common chord is zero.  Maximum length of the common chord = diameter of the smaller circle. Example : 38 If the common chord of the circles and subtend a right angle at the origin, then  is equal to (a) 4 (b) (c) (d) 8 Solution : (c) The common chord of given circles is i.e., ( 0) The pair of straight lines joining the origin to the points of intersection of and is . These lines are at right angles if , i.e., Example : 39 Which of the following is a point on the common chord of the circles and (a) (1, –2) (b) (1, 4) (c) (1, 2) (d) (1, – 4) Solution : (d) Given circles are, ….. (i) and ….. (ii) Equation of common chord is , and out of the four given points only point (1, – 4) satisfies it. Example : 40 If the circle bisects the circumference of the circle then a equals (a) 4 (b) – 4 (c) 16 (d) – 16 Solution : (c) The common chord of given circles is …..(i) Since, bisects the circumferences of the circle therefore (i) passes through the centre of second circle i.e. (1, – 3). 2 + 18 – 4 – a = 0  a = 16. 4.19 Angle of Intersection of Two Circles. The angle of intersection between two circles S = 0 and S' = 0 is defined as the angle between their tangents at their point of intersection. If are two circles with radii and d be the distance between their centres then the angle of intersection  between them is given by or (1) Condition of Orthogonality : If the angle of intersection of the two circles is a right angle , then such circles are called orthogonal circles and condition for their orthogonality is Note :  When the two circles intersect orthogonally then the length of tangent on one circle from the centre of other circle is equal to the radius of the other circle.  Equation of a circle intersecting the three circles orthogonally is Example : 41 A circle passes through the origin and has its centre on If it cuts orthogonally, then the equation of the circle is (a) (b) (c) (d) Solution : (c) Let the required circle be .......(i) This passes through (0, 0), therefore c = 0 The centre of (i) lies on y = x, hence g = f. Since (i) cuts the circle orthogonally, therefore and . Hence the required circle is . Example : 42 The centre of the circle, which cuts orthogonally each of the three circles and is (a) (3, 2) (b) (1, 2) (c) (2, 3) (d) (0, 2) Solution : (a) Let the circle is …..(i) Circle (i) cuts orthogonally each of the given three circles. Then according to condition …..(ii) …..(iii) …..(iv) On solving (ii), (iii) and (iv), . Therefore, the centre of the circle Example : 43 The locus of the centre of a circle which cuts orthogonally the circle and which touches is (a) (b) (c) (d) Solution : (d) Let the circle be …..(i) It cuts the circle orthogonally …..(ii) Circle (i) touches the line …..(iii) Eliminating c from (ii) and (iii), we get . Hence the locus of ( is 4.20 Family of Circles. (1) The equation of the family of circles passing through the point of intersection of two given circles S = 0 and S' = 0 is given as (where is a parameter, (2) The equation of the family of circles passing through the point of intersection of circle S = 0 and a line L = 0 is given as (where is a parameter) (3) The equation of the family of circles touching the circle S = 0 and the line L = 0 at their point of contact P is (where is a parameter) (4) The equation of a family of circles passing through two given points and can be written in the form  (where is a parameter) (5) The equation of family of circles, which touch at for any finite m is And if m is infinite, the family of circles is (where is a parameter) (6) Equation of the circles given in diagram is Example : 44 The equation of the circle through the points of intersection of and touching the line is (a) (b) (c) (d) Solution : (c) Family of circles is Centre is and radius Since it touches the line Hence Radius = perpendicular distance from centre to the line cannot be possible in case of circle, so Equation of circle is Example : 45 The equation of the circle through the points of intersection of the circles and with its centre on the line (a) (b) (c) (d) Solution : (b) Equation of any circle through the points of intersection of given circles is or, …..(i) Its centre lies on the line y = x . Then Substituting the value of in (i), we get the required equation as . 4.21 Radical Axis. The radical axis of two circles is the locus of a point which moves such that the lengths of the tangents drawn from it to the two circles are equal. Consider, …..(i) and …..(ii) Let be a point such that On squaring, Locus of is which is the required equation of radical axis of the given circles. Clearly this is a straight line. (1) Some properties of the radical axis (i) The radical axis and common chord are identical : Since the radical axis and common chord of two circles S = 0 and S' = 0 are the same straight line S – S' = 0, they are identical. The only difference is that the common chord exists only if the circles intersect in two real points, while the radical axis exists for all pair of circles irrespective of their position (Except when one circle is inside the other). (ii) The radical axis is perpendicular to the straight line which joins the centres of the circles : Consider, …..(i) and …..(ii) Since and are the centres of the circles (i) and (ii), then slope of the straight line (say) Equation of the radical axis is, Slope of radical axis is (say). Hence and radical axis are perpendicular to each other. (iii) The radical axis bisects common tangents of two circles : Let AB be the common tangent. If it meets the radical axis LM in M then MA and MB are two tangents to the circles. Hence MA = MB, since length of tangents are equal from any point on radical axis. Hence radical axis bisects the common tangent AB. If the two circles touch each other externally or internally then A and B coincide. In this case the common tangent itself becomes the radical axis. (iv) The radical axis of three circles taken in pairs are concurrent : Let the equations of three circles be …..(i) …..(ii) …..(iii) The radical axis of the above three circles taken in pairs are given by …..(iv) …..(v) .....(vi) Adding (iv), (v) and (vi), we find L.H.S. vanished identically. Thus the three lines are concurrent. (v) If two circles cut a third circle orthogonally, the radical axis of the two circles will 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. Let …..(i) …..(ii) …..(iii) Since (i) and (ii) both cut (iii) orthogonally, and Subtracting, we get …..(iv) Now radical axis of (i) and (ii) is Since it will pass through the centre of circle (iii) or …..(v) which is true by (iv). Note :  Radical axis need not always pass through the mid point of the line joining the centres of the two circles. 4.22 Radical Centre. The radical axes of three circles, taken in pairs, meet in a point, which is called their radical centre. Let the three circles be .....(i) , .....(ii) , .….(iii) Let OL, OM and ON be radical axes of the pair sets of circles and respectively. Equation of OL, OM and ON are respectively .....(iv) , .....(v), .....(vi) Let the straight lines (iv) and (v) i.e., OL and OM meet in O. The equation of any straight line passing through O is where  is any constant For , this equation become , which is by (vi), equation of ON. Thus the third radical axis also passes through the point where the straight lines (iv) and (v) meet. In the above figure O is the radical centre. (1) Properties of radical centre (i) Co-ordinates of radical centre can be found by solving the equations (ii) The radical centre of three circles described on the sides of a triangle as diameters is the orthocentre of the triangle : Draw perpendicular from A on BC. Therefore, the circles whose diameters are AB and AC passes through D and A. Hence AD is their radical axis. Similarly the radical axis of the circles on AB and BC as diameter is the perpendicular line from B on CA and radical axis of the circles on BC and CA as diameter is the perpendicular line from C on AB. Hence the radical axis of three circles meet in a point. This point I is radical centre but here radical centre is the point of intersection of altitudes i.e., AD, BE and CF. Hence radical centre = orthocentre. (iii) The radical centre of three given circles will be the centre of a fourth circle which cuts all the three circles orthogonally and the radius of the fourth circle is the length of tangent drawn from radical centre of the three given circles to any of these circles. Let the fourth circle be where (h, k) is centre of this circle and r be the radius. The centre of circle is the radical centre of the given circles and r is the length of tangent from (h, k) to any of the given three circles. Example : 46 The gradient of the radical axis of the circles and is (a) (b) (c) (d) Solution : (b) Equation of radical axis is , Radical axis is . Hence, gradient of radical axis = Example : 47 The equations of three circles are and The coordinates of the point from which the length of tangents drawn to each of the three circles is equal (a) (b) (2, 2) (c) (d) None of these Solution : (d) The required point is the radical centre of the three given circles Now, , and Solving these equations, we get . Hence the required point is . Example : 48 The equation of the circle, which passes through the point (2a, 0) and whose radical axis is with respect to the circle , will be (a) (b) (c) (d) Solution : (a) Equation of radical axis is Equation of required circle is It is passes through the point (2a, 0) , Equation of circle is 4.23 Co-Axial System of Circles. A system (or a family) of circles, every pair of which have the same radical axis, are called co-axial circles. (1) The equation of a system of co-axial circles, when the equation of the radical axis and of one circle of the system are and respectively, is is an arbitrary constant). (2) The equation of a co-axial system of circles, where the equation of any two circles of the system are and Respectively, is , or , Other form (3) The equation of a system of co-axial circles in the simplest form is , where g is variable and c is a constant. 4.24 Limiting Points. Limiting points of a system of co-axial circles are the centres of the point circles belonging to the family (Circles whose radii are zero are called point circles). (1) Limiting points of the co-axial system : Let the circle is …..(i) where g is variable and c is constant. Centre and the radius of (i) are and respectively. Let  Thus we get the two limiting points of the given co-axial system as and Clearly the above limiting points are real and distinct, real and coincident or imaginary according as c>, =, <0 (2) System of co-axial circles whose two limiting points are given : Let (, ) and be the two given limiting points. Then the corresponding point circles with zero radii are and or and The equation of co-axial system is where is a variable parameter. or Centre of this circle is …..(i) For limiting point, radius After solving, find . Substituting value of in (i), we get the limiting point of co-axial system. (3) Properties of limiting points (i) The limiting point of a system of co-axial circles are conjugate points with respect to any member of the system : Let the equation of any circle be ….(i) Limiting points of (i) are and . The polar of the point with respect to (i) is or or or and it clearly passes through the other limiting point . Similarly polar of the point with respect to (i) also passes through . Hence the limiting points of a system of co-axial circles are conjugate points. (ii) Every circle through the limiting points of a co-axial system is orthogonal to all circles of the system : Let the equation of any circle be …..(i) where g is a parameter and c is constant. Limiting points of (i) are and Now let …..(ii) be the equation of any circle. If it passes through the limiting points of (i), then and . Solving, we get and From (ii), .....(iii) where c is constant and is variable. Applying the condition of orthogonality on (i) and (iii) i.e., we find that i.e., 0 = 0 Hence condition is satisfied for all values of and . Example : 49 The point (2, 3) is a limiting point of a co-axial system of circles of which is a member. The coordinates of the other limiting point is given by (a) (b) (c) (d) Solution : (a) Equation of circle with (2, 3) as limiting point is or or represents the family of co-axial circles. . For limiting points r = 0 The limiting points are (2, 3) and or . Example : 50 In the co-axial system of circle where g is a parameter, if c > 0. Then the circles are (a) Orthogonal (b) Touching type (c) Intersecting type (d) Non intersecting type Solution : (d) The equation of a system of circle with its centre on the axis of x is . Any point on the radical axis is Putting, x = 0, If c is positive (c >0), we have no real point on radical axis, then circles are said to be non-intersecting type. 4.25 Image of the Circle by the Line Mirror. Let the circle be and line mirror . In this condition, radius of circle remains unchanged but centre changes. Let the centre of image circle be . Slope of slope of …..(i) and mid point of and lie on i.e., …..(ii) Solving (i) and (ii), we get Required image circle is , where Example : 51 The equation of the image of the circle by the line mirror is (a) (b) (c) (d) Solution : (d) The given circle and line are …..(i) and …..(ii) Centre and radius of circle (i) are (– 8, 12) and 5 respectively. Let the centre of the image circle be Then slope of slope of or or …..(iii) and mid point of i.e., lie on , then or …..(iv) Solving (iii) and (iv), we get Equation of the image circle is or 4.26 Some Important Results. (1) Concyclic points : If A, B, C, D are concyclic then , where be the centre of the circle. (2) Equation of the straight line joining two points  and  on the circle Required equation is (3) The point of intersection of the tangents at the point P () and Q () on the circle is (4) Maximum and Minimum distance of a point from the circle : Let any point and circle …..(i) The centre and radius of the circle are and respectively. The maximum and minimum distance from to the circle (i) are and (P inside or outside) where (5) Length of chord of contact is and area of the triangle formed by the pair of tangents and its chord of contact is Where R is the radius of the circle and L is the length of tangent from on S=0. Here . (6) Length of an external common tangent and internal common tangent to two circles is given by Length of external common tangent and length of internal common tangent [Applicable only when ] where d is the distane between the centres of two circles i.e., and r1 and r2 are the radii of two circles. (7) Family of circles circumscribing a triangle whose sides are given by and is given by provided coefficient of and coefficient of coefficient of Equation of the circle circumscribing the triangle formed by the lines where r = 1, 2, 3, is (8) Equation of circle circumscribing a quadrilateral whose sides in order are represented by the lines and is given by provided coefficient of coefficient of and coefficient of (9) Equation of the circle circumscribing the triangle PAB is where is the centre of the circle (10) Locus of mid point of a chord of a circle , which subtends an angle  at the centre is (11) The locus of mid point of chords of circle , which are making right angle at centre is (12) The locus of mid point of chords of circle , which are making right angle at origin is (13) The area of triangle , which is formed by co-ordinate axes and the tangent at a point of circle is (14) If a point is outside, on or inside the circle then number of tangents from the points is 2, 1 or none. (15) A variable point moves in such a way that sum of square of distances from the vertices of a triangle remains constant then its locus is a circle whose centre is the centroid of the triangle. (16) If the points where the line and meets the coordinate axes are concyclic then Example : 52 If are concylic points, then the value of is (a) 1 (b) – 1 (c) 0 (d) None of these Solution : (a) Let the equation of circle be Since the point lies on this circle Clearly its roots are and , = product of roots Example : 53 Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius r. If PS and RQ intersect at a point X on the circumference of the circle, then 2r equals (a) (b) (c) (d) Solution : (a) Also i.e.