∎ If S ≣ x^{2} + y^{2} + 2gx + 2fy + c = 0 and S’ ≣ x^{2} + y^{2} + 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 ≣ x^{2} + y^{2} + 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 − y_{1} = m(x − x_{1}) at (x_{1}, y_{1}) for any value of m is (x − x_{1})^{2} + (y − y_{1})^{2} + λ[(y − y_{1}) −m(x − x_{1})] = 0

If m is infinite, the equation is (x − x_{1})^{2} + (y − y_{1})^{2} + λ(x − x_{1}) = 0

Illustration : Find the equation of the circle described on the common chord of the circles

x^{2} + y^{2} − 4x − 5 = 0 and x^{2} + y^{2} + 8y + 7 = 0 as diameter.

Solution: Equation of the common chord is S_{1} − S_{2} = 0

=> x + 2y + 3 = 0

Equation of the circle through the two circles is S_{1} + λS_{2} = 0

=> x^{2} + y^{2} − 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 x^{2} + y^{2} − 2x + 4y + 1 = 0

Illustration : Tangents PQ and PR are drawn to the circle x^{2} + y^{2} = a^{2} from the point P(x_{1}, y_{1}). Prove that equation of the circum circle of ΔPQR is x^{2} + y^{2} − xx_{1} − yy_{1} = 0

Solution:

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

x^{2} + y^{2} −a^{2} = 0 ….(1)

equation of QR is xx_{1} + yy_{1} − a^{2} = 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(x_{1} , y_{1}).

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

x^{2} + y^{2} − a^{2} + λ (xx_{1} + yy_{1} − a^{2}) = 0 …(3)

it will pass through (x_{1}, y_{1}) if

x_{1}^{2} + y_{1}^{2} − a^{2} + λ (x_{1}^{2} + y_{1}^{2} − a^{2}) = 0

λ = −1 (since x_{1}^{2} + y_{1}^{2} ≠ a^{2})

Hence equation of circle is

(x^{2} + y^{2} − a^{2})−(xx_{1} + yy_{1} − a^{2}) = 0

Or x^{2} + y^{2} − xx_{1} − yy_{1} = 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 ≡ x^{2} + y^{2} + 2gx + 2fy +c = 0 and S’ ≡ x^{2} + y^{2} + 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 x^{2} + y^{2} – 2a_{1}x – 2b_{1}y + c_{1} = 0 and x^{2} + y^{2} – 2a_{2}x – 2b_{2}y + c_{2} = 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

x^{2} + y^{2} + 2gx + 2fy = 0, . . . . (1)

It cuts the given two circles orthogonally

=> – 2ga_{1} – 2fb_{1} = c_{1}

=> c_{1} + 2ga_{1} + 2fb_{1} = 0, . . . . . (2)

and – 2 g a_{2} – 2 f b_{2} = c_{2}

=> c_{2} + 2 g a_{2} + 2 f b_{2} = 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 x^{2} + y^{2} − 4x − 6y − 12 = 0 and x^{2} + y^{2} + 6x + 4y − 12 = 0 and cutting the circle x^{2} + y^{2} − 2x − 4 = 0 orthogonally.

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

x^{2} + y^{2} − 4x − 6y − 12 + λ(−10x −10y) = 0 (1)

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

x^{2} + y^{2} − 4x − 6y − 12 =0 and x^{2} + y^{2} + 6x + 4y – 12 = 0

Equation (1) can be re-arranged as

x^{2} + y^{2} − x(10λ + 4) −y(10λ + 6) − 12 = 0.

If cuts the circle x^{2} + y^{2} − 2x − 4 = 0 orthogonally.

Hence 2gg_{1} + 2ff_{1} = c + c_{1}

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

Hence the required circle is

x^{2} + y^{2} − 4x − 6y − 12 − 2(−10x − 10y) = 0

i.e., x^{2} + y^{2} + 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 x^{2} + y^{2} + x (2λ − 2) + y (3λ + 2) + 2 + λ = 0

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

Or x^{2} + y^{2} + 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 2x^{2} + 2y^{2} – 10x − 5y + 1 = 0

Exercise :

(i) The radical axis of the circles x^{2} + y^{2} + 2gx + 2fy + c = 0 and 2x^{2} + 2y^{2} + 3x + 8y + 2c = 0 touches the circle x^{2} + y^{2} + 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 x^{2} + y^{2} = a^{2} as diameter is x^{2}+ y^{2} − a^{2} − 2p (xcosα + ysinα − p) = 0.

(iv) Find the equation of the circle passing through the points of intersection of the circle x^{2} + y^{2} = 4a^{2}and x^{2} + y^{2} − 2x − 4y + 4 = 0 and touching the line x + 2y = 0

(v) If the circle C_{1} ; x^{2} + y^{2} = 16, intersects the circle C_{2} 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 C_{2}