Chapter-4-CIRCLE SYSTEM-01- (THEORY)-PART-I

4.1 Definition. A circle is defined as the locus of a point which moves in a plane such that its distance from a fixed point in that plane always remains the same i.e. constant. The fixed point is called the centre of the circle and the fixed distance is called the radius of the circle. Note :  If is the radius of a circle, the diameter is the maximum distance between any two points on the given circle.  The length of the curve or perimeter (also called circumference) of circle or .  The area of circle or .  Line joining any two points of a circle is called chord of circle.  Curved section between any two points of a circle is called arc of circle.  Angle subtended at the centre of a circle by any arc = arc/radius.  Angle subtended at the centre of a circle by an arc is double of angle subtended at the circumference of a circle. 4.2 Standard forms of Equation of a Circle. (1) General equation of a circle : The general equation of a circle is where g, f, c are constant. (i) Centre of the circle is (–g, –f). i.e., ( coefficient of x, coefficient of y) (ii) Radius of the circle is . Note :  The general equation of second degree represents a circle if and .  Locus of a point P represent a circle if its distance from two points A and B is not equal i.e. represent a circle if .  Discussion on nature of the circle :  If , then the radius of the circle will be real. Hence, in this case, it is possible to draw a circle on a plane.  If , then the radius of the circle will be zero. Such a circle is known as point circle.  If , then the radius of the circle will be an imaginary number. Hence, in this case, it is not possible to draw a circle.  Special features of the general equation of the circle : This equation has the following peculiarities :  It is a quadratic equation in x and y.  Here the co-efficient of the co-efficient of In working out problems it is advisable to keep the co-efficient of and as unity.  There is no term containing xy, i.e. the co-efficient of the term xy is zero.  This equation contains three arbitrary constants. If we want to find the equation of a circle of which neither the centre nor the radius is known, we take the equation in the above form and determine the values of the constants g, f, c for the circle in question from the given geometrical conditions.  Keeping in mind the above special features, we can say that the equation …..(i) also represents a circle. This equation can also be written as dividing by Hence, the centre and radius (2) Central form of equation of a circle : The equation of a circle having centre (h, k) and radius r is Note :  If the centre is origin, then the equation of the circle is  If r = 0, then circle is called point circle and its equation is (3) Concentric circle : Two circles having the same centre C (h, k) but different radii r1 and r2 respectively are called concentric circles. Thus the circles and are concentric circles. Therefore, the equations of concentric circles differ only in constant terms. (4) Circle on a given diameter : The equation of the circle drawn on the straight line joining two given points and as diameter is . Note :  If the coordinates of the end points of a diameter of a circle are given, we can also find the equation of the circle by finding the coordinates of the centre and radius. The centre is the mid-point of the diameter and radius is half of the length of the diameter. (5) Parametric coordinates (i) The parametric coordinates of any point on the circle are given by , In particular, co-ordinates of any point on the circle are , (ii) The parametric co-ordinates of any point on the circle are and , (6) Equation of a circle under given conditions: The general equation of circle, i.e., contains three independent constants g, f and c. Hence for determining the equation of a circle, three conditions are required. (i) The equation of the circle through three non-collinear points Let the equation of circle be …..(i) If three points lie on the circle (i), their co-ordinates must satisfy its equation. Hence solving equations …..(ii) …..(iii) …..(iv) g, f, c are obtained from (ii), (iii) and (iv). Then to find the circle (i). Alternative method (1) The equation of the circle through three non-collinear points is (2) From given three points taking any two as extremities of diameter of a circle S = 0 and equation of straight line passing through these two points is L = 0. Then required equation of circle is , where is a parameter which can be found out by putting third point in the equation. Note :  Cyclic quadrilateral : If all the four vertices of a quadrilateral lie on a circle, then the quadrilateral is called a cyclic quadrilateral. The four vertices are said to be concylic. 4.3 Equation of a Circle in Some special cases. (1) If centre of the circle is and it passes through origin then its equation is (2) If the circle touches x axis then its equation is (Four cases) (3) If the circle touches y axis then its equation is (Four cases) (4) If the circle touches both the axes then its equation is (Four cases) (5) If the circle touches x- axis at origin (Two cases) (6) If the circle touches y-axis at origin (Two cases) (7) If the circle passes through origin and cut intercepts of a and b on axes, the equation of circle is (Four cases) and centre is Note :  Circumcircle of a triangle : If we are given sides of a triangle, then first we should find vertices then we can find the equation of the circle using general form. Alternate : If equation of the sides are and , then equation of circle is , where and are the constant which can be found out by the conditions, coefficient of coefficient of and coefficient of xy = 0  If the triangle is right angled then its hypotenuse is the diameter of the circle. So using diameter form we can find the equation.  Circumcircle of a square or a rectangle : Diagonals of the square and rectangle will be diameters of the circumcircle. Hence finding the vertices of a diagonal, we can easily determine the required equation. Alternate : If sides of a quadrilateral are and . Then equation of circle is where is a constant which can be obtained by the condition of circle.  If a circle is passing through origin then constant term is absent i.e.  If the circle touches X-axis, then or  If the circle touches Y-axis, then or  If the circle touches both axes, then or g2 = f 2 = c Example: 1 A point P moves in such a way that the ratio of its distances from two coplanar points is always fixed number . Then its locus is (a) Straight line (b) Circle (c) Parabola (d) A pair of straight lines Solution: (b) Let two coplanar points are (0, 0) and (a, 0) and coordinates of point P is (x, y). Under given conditions, we get (where is any number and )   , which is equation of a circle. Example : 2 The lines and are the diameters of a circle of area 154 square units. The equation of the circle is (a) (b) (c) (d) Solution : (b) Centre of circle = Point of intersection of diameters, On solving equations, and , we get, Centre of circle = . Now area of circle = 154   r = 7 Hence, the equation of required circle is . Example : 3 The equation of a circle with origin as centre passing through the vertices of an equilateral triangle whose median is of length 3a is (a) (b) (c) (d) None of these Solution : (d) Since the triangle is equilateral, therefore centroid of the triangle is the same as the circumcentre and radius of the circum-circle (median) Centroid divides median in ratio of 2 : 1] Hence, the equation of the circum-circle whose centre is (0, 0) and radius 2a is Example : 4 A circle of radius 5 units touches both the axes and lies in first quadrant. If the circle makes one complete roll on x-axis along the positive direction of x-axis, then its equation in the new position is (a) (b) (c) (d) None of these Solution : (d) The x-coordinate of the new position of the circle is 5 + circumferrence of the first circle The y-coordinate is 5 and the radius is also 5. Hence, the equation of the circle in the new position is Example : 5 The abscissae of A and B are the roots of the equation and their ordinates are the roots of the equation The equation of the circle with AB as diameter is (a) (b) (c) (d) None of these Solution : (a) Let and be roots of and respectively. Then, and The equation of the circle with and as the end points of diameter is ; Example : 6 The equation of a circle of radius 1 touching the circles is (a) (b) (c) (d) Solution : (b,c) The given circles are and From the figure, the centres of the required circles will be and . The equations of the circles are Example : 7 If the line is a diameter of the circle then b = (a) 3 (b) – 5 (c) – 1 (d) 5 Solution : (d) Here the centre of circle (3, –1) must lie on the line Therefore, Example : 8 The centre of the circle is (a) (2, 3) (b) (– 2, 3) (c) (– 2, – 3) (d) (2, – 3) Solution : (b) Let r cos  = x and r sin  = y Squaring and adding, we get . Putting these values in given equation, Hence, centre of the circle = (–2, 3) Example : 9 The number of integral values of for which is the equation of a circle whose radius cannot exceed 5, is (a) 14 (b) 18 (c) 16 (d) None of these Solution : (c) Centre of circle ; Radius of circle , (Nearly). ................,7, 8. Hence number of integral values of is 16. Example : 10 Let be the equation of a circle. If has equal roots and has roots then the centre of the circle is (a) (b) (c) (d) None of these Solution : (b) Now, and its roots are 2, 2. i.e. and its roots are i.e., . Hence, centre of the circle . Example : 11 If the lines and are tangents to a circle, then the radius of the circle is (a) 3/2 (b) 3/4 (c) 1/10 (d) 1/20 Solution : (b) Since both tangents are parallel to each other. The diameter of the circle is perpendicular distance between the parallel lines (tangents) and and so it is equal to . Hence radius of circle is . Alternative method : Perpendicular distance = , i.e., Diameter = Hence radius of circle is . 4.4 Intercepts on the Axes. The lengths of intercepts made by the circle with X and Y axes are and respectively. Let the equation of circle be ......(i) Length of intercepts on x-axis and y-axis are and respectively. The circle intersects the x-axis, when y = 0, then . Since the circle intersects the x-axis at and . Then As the circle intersects the y-axis, when x = 0, then Since the circle intersects the y-axis at C (0, y1) and D (0, y2), then . Note :  If then the roots of the equation are real and distinct, so the circle meets the x-axis in two real and distinct points and the length of the intercept on x-axis is  If then the roots of the equation are real and equal, so the circle touches x-axis and the intercept on x-axis is zero.  If then the roots of the equation are imaginary, so the circle does not meet x-axis in real points.  Similarly, the circle cuts the y-axis in real and distinct points, touches or does not meet in real points according as . 4.5 Position of a point with respect to a Circle. A point lies outside, on or inside a circle according as is positive, zero or negative i.e., Point is outside the circle. Point is on the circle. Point is inside the circle. (1) The least and greatest distance of a point from a circle : Let S = 0 be a circle and be a point. If the diameter of the circle is passing through the circle at P and Q, then least distance ; greatest distance where 'r' is the radius and C is the centre of the circle. Example : 12 The number of points with integral coordinates that are interior to the circle is (a) 43 (b) 49 (c) 45 (d) 51 Solution : (c) The number of points is equal to the number of integral solutions (x, y) such that . So, x, y are integers such that satisfying the inequation . The number of selections of values of x is 7, namely –3, –2, –1, 0, 1, 2, 3. The same is true for y. So the number of ordered pairs (x, y) is 7×7. But (3, 3), (3, –3), (–3, 3), (–3, –3) are rejected because they do not satisfy the inequation So the number of points is 45. Example : 13 The range of values of a for which the point (a, 4) is outside the circles and is (a) (b) (– 8, – 2) (c) (d) None of these Solution : (a) For circle, ; or …..(i) For circle, ;  …..(ii) Taking common values from (i) and (ii), . 4.6 Intersection of a Line and a Circle. Let the equation of the circle be …..(i) and the equation of the line be …..(ii) From (i) and (ii), or …..(iii) Case I: When points of intersection are real and distinct. In this case (iii) has two distinct roots. or or or length of perpendicular from (0, 0) to a > length of perpendicular from (0, 0) to Thus, a line intersects a given circle at two distinct points if radius of circle is greater than the length of perpendicular from centre of the circle to the line. Case II: When the points of intersection are coincident in this case (iii) has two equal roots. or a = length of perpendicular from the point (0, 0) to . Thus, a line touches the circle if radius of circle is equal to the length of perpendicular from centre of the circle to the line. Case III: When the points of intersection are imaginary. In this case (iii) has imaginary roots.  , or length of perpendicular from (0, 0) to a < length of perpendicular from (0, 0) to Thus, a line does not intersect a circle if the radius of circle is less than the length of perpendicular from centre of the circle to the line. (1) The length of the intercept cut off from a line by a circle : The length of the intercept cut off from the line by the circle is (2) Condition of tangency : A line L = 0 touches the circle S = 0, if length of perpendicular drawn from the centre of the circle to the line is equal to radius of the circle i.e. p = r. This is the condition of tangency for the line L = 0. Circle will touch the line if Again, (i) If line will meet the circle at real and different points. (ii) If line will touch the circle. (iii) If line will meet circle at two imaginary points. Example : 14 If the straight line is outside the circle then (a) m > 3 (b) m < 3 (c) | m | > 3 (d) | m | < 3 Solution : (d) If the straight line is outside the given circle then perpendicular distance of line from centre of circle > radius of circle Example : 15 If the chord of the circle subtends an angle of measure 45o at the major segment of the circle then value of m is (a) 2 (b) – 2 (c) 1 (d) None of these Solution : (c) Given circle is C (0,0) and radius = 1 and chord is ; CP = Perpendicular distance from (0, 0) to chord (CR = radius =1)  4.7 Tangent to a Circle at a given Point. The limiting position of the line PQ, when Q moves towards P and ultimately coincides with P, is called the tangent to the circle at the point P. The point P is called the point of contact. (1) Point form (i) The equation of tangent at (x1, y1) to circle is (ii) The equation of tangent at to circle is Note :  For equation of tangent of circle at , substitute for for for for y and for xy and keep the constant as such.  This method of tangent at is applied any conics of second degree. i.e., equation of tangent of at is (2) Parametric form : Since parametric co-ordinates of circle is then equation of tangent at is or . (3) Slope form : Let is the tangent of the circle . Length of perpendicular from centre of circle (0, 0) on line = radius of circle Substituting this value of c in we get . Which are the required equations of tangents. Note :  The reason why there are two equations is that there are two tangents, both are parallel and at the ends of a diameter.  The line is a tangent to the circle if and only if  The condition that the line touches the circle is  Equation of tangent to the circle in terms of slope is (4) Point of contact : If circle be and tangent in terms of slope be , Solving and simultaneously, we get and Thus, the co-ordinates of the points of contact are Alternative method : Let point of contact be then tangent at of is . Since and are identical , and Thus, the co-ordinates of the points of contact are Note :  If the line y = mx + c is the tangent to the circle then point of contact is given by  If the line ax+by+c = 0 is the tangent to the circle x2+y2=r2 then point of contact is given by Example : 16 The equations to the tangents to the circle which are parallel to the straight line 4x+3y+5=0, are (a) (b) (c) (d) Solution : (c) Let equation of tangent be then  and – 31 Hence the equations of tangents are and Example : 17 The equations of any tangents to the circle is (a) (b) (c) (d) None of these Solution : (a) Equation of circle is . As any tangent to is given by Any tangent to the given circle will be Example : 18 If a circle, whose centre is (–1, 1) touches the straight line then the coordinates of the point of contact are [MP PET 1998] (a) (b) (c) (2, –7) (d) (– 2, – 5) Solution : (b) Let point of contact be This point lies on the given line ,  ….. (i) Gradient of , Gradient of Both are perpendicular, ….. (ii) On solving the equation (i) and (ii), 4.8 Length of Tangent. From any point, say two tangents can be drawn to a circle which are real, coincident or imaginary according as P lies outside, on or inside the circle. Le PQ and PR be two tangents drawn from to the circle Then PQ =PR is called the length of tangent drawn from point P and is given by PQ = PR 4.9 Pair of Tangents. From a given point two tangents PQ and PR can be drawn to the circle Their combined equation is Where is the equation of circle, is the equation of tangent at and S1 is obtained by replacing x by x1 and y by y1 in S. 4.10 Power of Point with respect to a Circle. Let be a point outside the circle and PAB and PCD drawn two secants. The power of with respect to is equal to PA . PB which is Power remains constant for the circle i.e., independent of A and B. square of the length of tangent. Note :  If P is outside, inside or on the circle then PA . PB is +ve, –ve or zero respectively. Important Tips  The length of the tangent drawn from any point on the circle to the circle is .  If two tangents drawn from the origin to the circle are perpendicular to each other, then  If the tangent to the circle at the point (a, b) meets the coordinate axes at the points A and B and O is the origin, then the area of the triangle OAB is .  If  is the angle subtended at by the circle then  The angle between the tangents from to the circle is .  If OA and OB are the tangents from the origin to the circle and C is the centre of the circle, then the area of the quadrilateral OACB is Example : 19 If the distances from the origin to the centres of three circles are in G.P. then the lengths of the tangents drawn to them from any point on the circle are in (a) A.P. (b) G.P. (c) H.P. (d) None of these Solution : (b) The centres of the given circles are The distances from the origin to the centres are It is given that Let be any point on the circle , then, Now, = length of the tangent from (h, k) to = Therefore, . Hence, are in G.P. Example : 20 From a point on the circle , two tangents are drawn to the circle The angle between them is (a)  (b) (c) (d) None of these Solution : (c) Let any point on the circle be and Now; PQ = length of tangent from P on the circle Radius of the circle , ; Angle between tangents Alternative Method : We know that, angle between the tangent from to the circle is . Let point on the circle be Angle between tangent = Example : 21 Two tangents to the circle at the points A and B meet at P (– 4, 0). The area of quadrilateral PAOB, where O is the origin, is (a) 4 (b) (c) (d) None of these Solution : (c) Clearly, , So area Area (quadrilateral PAOB) Trick : Area of quadrilateral Example : 22 The angle between a pair of tangents drawn from a point P to the circle is The equation of the locus of the point P is (a) (b) (c) (d) Solution : (d) The centre of the circle is and its radius is Let P (h, k) be any point on the locus. The Also i.e. triangle APC is a right angle triangle. Thus or Thus the required equation of the locus is . 4.11 Normal to a Circle at a given Point. The normal of a circle at any point is a straight line, which is perpendicular to the tangent at the point and always passes through the centre of the circle. (1) Equation of normal: The equation of normal to the circle at any point is or Note :  The equation of normal to the circle at any point is or  The equation of any normal to the circle is where m is the slope of normal.  The equation of any normal to the circle is where m is the slope of normal.  If the line is a normal to the circle with radius r and centre at (a, b) then . (2) Parametric form : Since parametric co-ordinates of circle is . Equation of normal at is or or or where , which is slope form of normal. Example : 23 The line is a normal to the circle if (a) (b) (c) (d) Solution : (a) Since normal always passes through centre of circle, therefore must lie on . Hence, 4.12 Chord of Contact of Tangents. (1) Chord of contact : The chord joining the points of contact of the two tangents to a conic drawn from a given point, outside it, is called the chord of contact of tangents. (2) Equation of chord of contact : The equation of the chord of contact of tangents drawn from a point to the circle is Equation of chord of contact at to the circle is It is clear from the above that the equation to the chord of contact coincides with the equation of the tangent, if point lies on the circle. The length of chord of contact ; (p being length of perpendicular from centre to the chord) Area of is given by . (3) Equation of the chord bisected at a given point : The equation of the chord of the circle bisected at the point is given by i.e. . Example : 24 Tangents are drawn from any point on the circle to the circle . If the chord of contact touches the circle then (a) a, b, c are in A.P. (b) a, b, c are in G.P. (c) a, b, c are in H.P. (d) a, c, b are in G.P. Solution : (b) Chord of contact of any point on 1st circle with respect to 2nd circle is This chord touches the circle Hence, Radius = Perpendicular distance of chord from centre. .Hence a,b,c are in G.P. Example : 25 The area of the triangle formed by the tangents from the point (4, 3) to the circle and the line joining their points of contact is (a) (b) (c) (d) None of these Solution : (b) The equation of the chord of contact of tangents drawn from P (4, 3) to is The equation of OP is Now, OM = (length of the perpendicular from (0, 0) on Now, .So, Area of Example : 26 The locus of the middle points of those chords of the circle which subtend a right angle at the origin is (a) (b) (c) (d) Solution : (c) Let the mid-point of chord is (h, k). Also radius of circle is 2. Therefore Hence locus is Example : 27 If two distinct chords, drawn from the point (p, q) on the circle (where p, q  0) are bisected by the x-axis, then (a) (b) (c) (d) Solution : (d) Let (h, 0) be a point on x-axis, then the equation of chord whose mid-point is (h, 0) will be . This passes through (p, q), hence ; h is real, hence 4.13 Director Circle. The locus of the point of intersection of two perpendicular tangents to a circle is called the Director circle. Let the circle be , then equation of the pair of tangents to a circle from a point is . If this represents a pair of perpendicular lines, coefficient of coefficient of i.e. Hence the equation of director circle is . Obviously director circle is a concentric circle whose radius is times the radius of the given circle. Note :  Director circle of circle is . 4.14 Diameter of a Circle. The locus of the middle points of a system of parallel chords of a circle is called a diameter of the circle. The equation of the diameter bisecting parallel chords (c is a parameter) of the circle is Note :  The diameter corresponding to a system of parallel chords of a circle always passes through the centre of the circle and is perpendicular to the parallel chords. Example : 28 A foot of the normal from the point (4, 3) to a circle is (2, 1) and a diameter of the circle has the equation Then the equation of the circle is (a) (b) (c) (d) None of these. Solution : (b) The line joining (4, 3) and (2, 1) is also along a diameter. So, the centre is the intersection of the diameters and . Solving these, the centre = (1, 0) Radius = Distance between (1, 0) and (2, 1) = Equation of circle Example : 29 The diameter of the circle corresponding to a system of chords parallel to the line (a) (b) (c) (d) None of these Solution : (c) The centre of the given circle is (2, –1) the equation of the line perpendicular to chord is Since the line passes through the point (2, –1) therefore k = – 3. The equation of diameter is 4.15 Pole and Polar. Let be any point inside or outside the circle. Draw chords AB and A' B' passing through P. If tangents to the circle at A and B meet at Q (h, k), then locus of Q is called the polar of P with respect to circle and P is called the pole and if tangents to the circle at A' and B' meet at Q', then the straight line QQ' is polar with P as its pole. If circle be then AB is the chord of contact of Q (h, k), is its equation. But lies on AB, . Hence, locus of Q (h, k) is , which is polar of with respect to the circle . (1) Coordinates of pole of a line : The pole of the line with respect to the circle . Let pole be then equation of polar with respect to the circle is , which is same as Then , and . Hence, the required pole is . (2) Properties of pole and polar (i) If the polar of w.r.t. a circle passes through then the polar of Q will pass through P and such points are said to be conjugate points. (ii) If the pole of the line w.r.t. a circle lies on another line then the pole of the second line will lie on the first and such lines are said to be conjugate lines. (iii) The distance of any two points and from the centre of a circle is proportional to the distance of each from the polar of the other. (iv) If O be the centre of a circle and P any point, then OP is perpendicular to the polar of P. (v) If O be the centre of a circle and P any point, then if OP (produced, if necessary) meet the polar of P in Q, then OP . OQ = (radius)2. Note :  Equation of polar is like as equation of tangent i.e., T = 0 (but point different)  Equation of polar of the circle with respect to is  If the point P is outside the circle then equation of polar and chord of contact will coincide. In this case the polar cuts the circle at two points.  If the point P is on the circle then equation of polar, chord of contact and tangent at P will coincide. So in this case the polar touches the circle.  If the point P is inside the circle (not its centre) then only its polar will exist. In this case the polar is outside the circle. The polar of the centre lies at infinity.  If a triangle is like that its each vertex is a pole of opposite side with respect to a circle then it is called self conjugate triangle. Example : 30 The polar of the point with respect to circle is (a) (b) (c) (d) Solution : (b) The polar of the point is . Example : 31 The pole of the straight line with respect to circle is (a) (3, 1) (b) (1, 3) (c) (3, – 1) (d) (– 3, 1) Solution : (c) Equation of given circle is  Put and we get the equation of circle and the line Hence pole . But, and , hence the pole is (3, – 1). 4.16 Two Circles touching each other. (1) When two circles touch each other externally : Then distance between their centres = Sum of their radii i.e., In such cases, the point of contact P divides the line joining C1 and C2 internally in the ratio If and , then co-ordinate of P is (2) When two circles touch each other internally : Then distance between their centres = Difference of their radii i.e., In such cases, the point of contact P divides the line joining C1 and C2 externally in the ratio If and , then co-ordinate of P is 4.17 Common Tangents to Two circles. Different cases of intersection of two circles : Let the two circles be …..(i) and …..(ii) with centres and and radii r1 and r2 respectively. Then following cases may arise : Case I : When i.e., the distance between the centres is greater than the sum of radii. In this case four common tangents can be drawn to the two circles, in which two are direct common tangents and the other two are transverse common tangents. Case II : When i.e., the distance between the centres is equal to the sum of radii. In this case two direct common tangents are real and distinct while the transverse tangents are coincident. Case III : When i.e., the distance between the centres is less than sum of radii. In this case two direct common tangents are real and distinct while the transverse tangents are imaginary. Case IV : When i.e., the distance between the centres is equal to the difference of the radii. In this case two tangents are real and coincident while the other two tangents are imaginary. Case V : When i.e., the distance between the centres is less than the difference of the radii. In this case, all the four common tangents are imaginary. Note :  Points of intersection of common tangents : The points and (points of intersection of indirect and direct common tangents) divides internally and externally in the ratio respectively.  Equation of the common tangents at point of contact is  If the circle and touch each other, then .