# Equation of Family of Circles

∎ If S ≣ x2 + y2 + 2gx + 2fy + c = 0 and S’ ≣ x2 + y2 + 2g’x + 2f’y + c’ = 0 are two intersecting circles, then S + λS’ = 0, λ ≠ −1 , is the equation of the family of circles passing through the points of intersection of S = 0 and S’ = 0

∎ If S ≣ x2 + y2 + 2gx + 2fy + c = 0 is a circle which is intersected by the straight line μ ≣ ax + by + c = 0 at two real and distinct points, then S + λμ = 0 is the equation of the family of circles passing through the points of intersection of S = 0 and μ = 0.

If μ = 0 touches S = 0 at P, then S + λμ = 0 is the equation of the family of circles, each touching μ = 0 at P.

∎ 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 : Find the equation of the circle described on the common chord of the circles
x2 + y2 − 4x − 5 = 0 and x2 + y2 + 8y + 7 = 0 as diameter.

Solution: Equation of the common chord is S1 − S2 = 0

=> x + 2y + 3 = 0

Equation of the circle through the two circles is S1 + λS2 = 0

=> x2 + y2 − 4x/(1+λ) + 8λy/(1+λ) + (7λ − 5)/(1 + λ) = 0

Its centre $\large \frac{2}{1+\lambda} , \frac{-4}{1+\lambda}$

lies on x + 2y + 3 = 0

$\large \frac{2}{1+\lambda} + 2(\frac{-4}{1+\lambda}) + 3 = 0$

=> 2 − 8λ + 3 + 3λ = 0

=> λ = 1.

Hence the required circle is x2 + y2 − 2x + 4y + 1 = 0

Illustration : Tangents PQ and PR are drawn to the circle x2 + y2 = a2 from the point P(x1, y1). Prove that equation of the circum circle of ΔPQR is x2 + y2 − xx1 − yy1 = 0

Solution: QR is the chord of contact of the tangents to the circle

x2 + y2 −a2 = 0 ….(1)

equation of QR is xx1 + yy1 − a2 = 0 …..(2)

The circumcircle of ΔPQR is a circle passing through the intersection of the circle (1) and the line (2) and the point P(x1 , y1).

Circle through the intersection of (1) and (2) is

x2 + y2 − a2 + λ (xx1 + yy1 − a2) = 0 …(3)

it will pass through (x1, y1) if

x12 + y12 − a2 + λ (x12 + y12 − a2) = 0

λ = −1 (since x12 + y12 ≠ a2)

Hence equation of circle is

(x2 + y2 − a2)−(xx1 + yy1 − a2) = 0

Or x2 + y2 − xx1 − yy1 = 0

##### Angle of Intersection of two Circles The angle between the two circles is the angle between their tangents at their point of intersection. The angle of intersection θ of two circles S ≡ x2 + y2 + 2gx + 2fy +c = 0 and S’ ≡ x2 + y2 + 2g’x + 2f’y +c’ = 0 is given by

$\large cos\theta = \pm ( \frac{2 g g’ + 2 f f’-c – c’}{2\sqrt{g^2 + f^2 -c} \sqrt{g’^2 + f’^2 -c’}} )$

##### Orthogonal Intersection of Two Circles:

The two circles are said to intersect orthogonally if the angle of intersection of the circles i.e., the angle between their tangents at the point of intersection, is 90° .

The condition for the two circles S = 0 and S’1= 0 to cut each other orthogonally is 2gg’ + 2ff’ = c + c’

Illustration : Show that the circle passing through the origin and cutting the circles x2 + y2 – 2a1x – 2b1y + c1 = 0 and x2 + y2 – 2a2x – 2b2y + c2 = 0 orthogonally is

$\large \left| \begin{array}{ccc} x^2 + y^2 & x & y \\ c_1 & a_1 & b_1 \\ c_2 & a_2 & b_2 \end{array} \right| = 0$

Solution: Let the equation of the circle passing through the origin be

x2 + y2 + 2gx + 2fy = 0, . . . . (1)

It cuts the given two circles orthogonally

=> – 2ga1 – 2fb1 = c1

=> c1 + 2ga1 + 2fb1 = 0, . . . . . (2)

and – 2 g a2 – 2 f b2 = c2

=> c2 + 2 g a2 + 2 f b2 = 0. . . . . . . (3)

Eliminating 2f and 2g from (1), (2) and (3), we get

$\large \left| \begin{array}{ccc} x^2 + y^2 & x & y \\ c_1 & a_1 & b_1 \\ c_2 & a_2 & b_2 \end{array} \right| = 0$

Illustration : Find the equation of the circle through the points of intersection of the circles x2 + y2 − 4x − 6y − 12 = 0 and x2 + y2 + 6x + 4y − 12 = 0 and cutting the circle x2 + y2 − 2x − 4 = 0 orthogonally.

Solution: The equation of the circle through the intersection of the given circles is

x2 + y2 − 4x − 6y − 12 + λ(−10x −10y) = 0 (1)

Where ( − 10x − 10y = 0) is the equation of radical axis for the circle

x2 + y2 − 4x − 6y − 12 =0 and x2 + y2 + 6x + 4y – 12 = 0

Equation (1) can be re-arranged as

x2 + y2 − x(10λ + 4) −y(10λ + 6) − 12 = 0.

If cuts the circle x2 + y2 − 2x − 4 = 0 orthogonally.

Hence 2gg1 + 2ff1 = c + c1

=> 2(5λ + 2)(1) + 2(5λ + 3) (0) = −12 − 4 => λ = −2

Hence the required circle is

x2 + y2 − 4x − 6y − 12 − 2(−10x − 10y) = 0

i.e., x2 + y2 + 16x + 14y – 12 = 0

Illustration : A circle touches the line 2x + 3y + 1 = 0 at the point (1, −1) and is orthogonal to the circle which has the line segment having end points (0, −1) and (−2, 3) as the diameter.

Solution: Let the circle with tangent 2x + 3y + 1 = 0 at (1, − 1) be

(x − 1)2 + (y + 1)2 + λ (2x + 3y + 1) = 0

or x2 + y2 + x (2λ − 2) + y (3λ + 2) + 2 + λ = 0

It is orthogonal to x(x + 2) + (y + 1)(y − 3) = 0

Or x2 + y2 + 2x − 2y − 3 = 0

so that, $\large \frac{2(2\lambda-2)}{2} . \frac{2}{2}+ \frac{2(3\lambda + 2)}{2} \frac{-2}{2} = 2 + \lambda – 3$

=> λ = −3/2

Hence the required circle is 2x2 + 2y2 – 10x − 5y + 1 = 0

Exercise :

(i) The radical axis of the circles x2 + y2 + 2gx + 2fy + c = 0 and 2x2 + 2y2 + 3x + 8y + 2c = 0 touches the circle x2 + y2 + 2x − 2y + 1 = 0. Show that either g = 3/4 or f = 2

(ii) The equation of radical axis of two circles is x + y = 1. One of these 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 equations of these circles.

(iii) Show that the circle on the chord xcosα + ysinα − p = 0 of the circle x2 + y2 = a2 as diameter is x2+ y2 − a2 − 2p (xcosα + ysinα − p) = 0.

(iv) Find the equation of the circle passing through the points of intersection of the circle x2 + y2 = 4a2and x2 + y2 − 2x − 4y + 4 = 0 and touching the line x + 2y = 0

(v) If the circle C1 ; x2 + y2 = 16, intersects the circle C2 of radius 5 in such a manner that the common chord is of maximum length, and has slope equal to 3/4, then find the coordinates of the centre of C2