The radical axis of two circles is the locus of a point which moves in such a way that the tangent segments drawn from it to the two circles are of equal length.

Equation to the Radical Axis:

In general 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

where S ≣ x^{2} + y^{2} + 2gx + 2fy + c = 0 and S’ ≣ x^{2} + y^{2} + 2g’x + 2f’y + c’ = 0

∎ If S = 0 and S’ = 0 intersect in two real and distinct points then S – S’ = 0 is the equation of the common chord of the two circles.

∎ If S’ = 0 and S = 0 touch each other (internally or externally), then S – S’ = 0 is the equation of the common tangent to the two circles at the point of contact.

Note:

∎ To find the equation of the radical axis of two circles, first make the coefficients of x^{2} and y^{2} in the equation of the two circles equal to unity.

__Length of the common chord:__

From figure, PQ = 2PR = 2√(AP)^{2}-(AR)^{2} , where AP is the radius of the circle, S = 0 and AR is the length of the perpendicular from A to the common chord PQ.

Notes:

∎ The length of the common chord of the two circles becomes maximum when it is a diameter of the smaller one between them.

∎ If the length of the common chord is zero, then the two circles touch each other and the common chord becomes the common tangent to the two circles at the common point of contact.

Properties of Radical Axis

∎ The radical axis of two circles is perpendicular to the line joining their centres.♦

∎ Radical centre: 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 S_{1} = S_{2} = S_{3} = 0.

∎ The radical centre of three circles described on the sides of a triangle as diameters is the orthocentre of the triangle.

∎ If two circles cut a third circle orthogonally, then the radical axis of the two circles pass through the centre of the third circle.

∎ The radical axis of the two circles will bisect their common tangents.

Example : Find the coordinates of the points at which the circles x^{2} + y^{2} – 4x – 2y = 4 and x^{2} + y^{2} − 12x − 8y = 12 touch each other. Find the coordinates of the point of contact and equation of the common tangent at the point of contact.

Solution: The given circles are

S_{1} ≣ x^{2} + y^{2} − 4x − 2y − 4 = 0 and S_{2} ≣ x^{2} + y^{2} − 12x − 8y − 12 = 0

Equation of the common tangent is S_{1} − S_{2} = 0 i.e. 8x + 6y + 8 = 0

4x + 3y + 4 = 0 . . . (1)

Let the point of contact be (x_{1}, y_{1})

Equation of the tangent at (x_{1}, y_{1}) is ,

xx_{1} + yy_{1} − 2(x + x_{1}) − (y + y_{1}) − 4 = 0

or (x_{1} − 2)x + (y_{1} − 1)y − 2x_{1} − y_{1} − 4 = 0 . . . (2)

Comparing (1) and (2), we get

$\large \frac{x_1 -2}{4} = \frac{y_1 -1}{3} = \frac{2x_1 + y_1 + 4}{-4} = k (say)$

x_{1} = 4k + 2, y_{1} = 3k + 1 , 2x_{1} + y_{1} + 4 = − 4k

2(4k + 2) + 3k + 1 + 4 = − 4k

or 8k + 4 + 3k + 5 = − 4k

15k = −9 => k = −9/15 = −3/5

x_{1} = 4 x(−3/5)+2 , y_{1} = 3 x (−3/5)+ 1

Point of contact $\large (\frac{-2}{5} , \frac{-4}{5}) $

Example : (a) If two circles cut a third circle orthogonally, prove that their common chord will pass through the centre of the third circle.

(b) A and B are two fixed points and P moves such that PA = n PB ; where n ≠ 1. Show that locus of P is a circle and for different values of n all the circles have a common radical axis.

Solution: (a) Let us take the equation of the two circles as

x^{2} + y^{2} + 2g_{1}x + 2f_{1}y + c_{1}= 0 ….(1)

x^{2} + y^{2} + 2g_{2}x + 2f_{2}y + c_{2} = 0. ….(2)

Let the third circle be

x^{2}+ y^{2} + 2gx + 2fy + c = 0. …(3)

Since (1) and (2) cut the third circle orthogonally

2g_{1}g + 2f_{1}f = c_{1} + c …(4)

and 2g_{2}g + 2f_{2}f = c_{2} + c ….(5)

(5) − (4)

2 (g_{2} − g_{1})g + 2 (f_{2} − f_{1}f = c_{2} − c_{1} …(6)

Also radical axis of (1) and (2) is

2 (g_{1} − g_{2})x + (f_{1} − f_{2})y = c_{2} − c_{1} ….(7)

Put (− g , − f) in (7), we get

2 (g_{2} − g_{1})g + 2 (f_{2} − f_{1})f = c_{2} − c_{1}

Hence common chord of (1) and (2) pass through the centre of the third circle.

(b) Let A ≣ (a, 0), B ≣ (−a, 0) and P(h , k)

so PA = n PB

( h − a)^{2} + k^{2} = n^{2} [(h + a)^{2} + k^{2}]

( 1 − n^{2})h^{2} + (1 − n^{2})k^{2} − 2ah (1 + n^{2}) + (1 − n^{2})a^{2} = 0

$\large h^2 + k^2 – 2 a h (\frac{1+n^2}{1-n^2}) + a^2 = 0 $

Hence locus of P is

$\large x^2 + y^2 – 2 a x (\frac{1+n^2}{1-n^2}) + a^2 = 0 $ , which is a circle for different values of n.

Let n_{1} and n_{2} are two different values of n so their radical axis is x = 0 i.e. y-axis. Hence for different values of n the circles have a common radical axis.