Posted on - 02-02-2017

IIT JEE

The radical axis of two circles is the locus of a point from which the tangent segments 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 º x^{2} + y^{2} + 2gx
+ 2fy + c=0 and

S' º x^{2} + y^{2} + 2g'x + 2f'y + c' =0.

· If S = 0 and S' = 0 intersect in 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, then S - S' = 0 is the equation of the common tangent to the two circles at the point of contact.

Prove
that the circle x^{2} + y^{2} – 6x – 4y + 9 = 0 bisects the
circumference of the circle x^{2} + y^{2} – 8x – 6y + 23 = 0.

The given circles are

S_{1} º
x^{2} + y^{2} – 6x – 4y + 9 = 0 ………..(1)
.

and S_{2} º
x^{2} + y^{2} – 8x – 6y + 23 = 0. …….….(2).

Equation of the common
chord of circles (1) and (2) which is also the radical axis of the circles S_{1}
and S_{2} is

S_{1} – S_{2}
= 0 or, 2x + 2y – 14 = 0

or, x + y – 7 = 0 ………..(3).

Centre of the circle S_{2}
is (4, 3). Clearly, line (3) passes through the point (4, 3) and hence line
(3) is the equation of the diameter of the circle (2). Hence circle (1) bisects
the circumference of circle (2).

If two circles cut a third circle orthogonaly, prove that their common chord will pass through the centre of the third circle. .

Let us take the equation of the two circles as

x^{2} + y^{2} + 2l_{1}x + a = 0 ….(1) .

x^{2} + y^{2} + 2l_{2}x + a = 0 ….(2) .

We can select axes suitable (the line of centres as x–axis and the point midway between the centres as origin) to get the above form of equation. .

Let the third circle be x^{2}
+ y^{2} + 2gx + 2fy + c = 0 ….(3) .

The circle (1) and (3) cut
orthogonally 2l_{1}g = a + c ….(4) .

The circles (2) and (3) cut orthogonally

2l_{2} g = a + c …(5)

From (4) and (5), 2g(l_{1} – l_{2}) = 0 but l_{1} > l_{2}

\ g = 0

Hence centre of the third circle (0, –f)

The common chord of (1) and (2) has
the equation S_{1} – S_{2} = 0

i.e. (x^{2} + y^{2} +
2l_{1}x
+ a) – (x^{2} + y^{2} + 2l_{2}x + a) = 0 .

or 2(l_{1} – l_{2}) x = 0 \
x = 0

\ (0, –f) satisfies the equation x = 0.