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Circle and Radical Axis for IIT JEE and other Engineering Exams

Posted on - 02-02-2017

JEE Math Circle

IIT JEE

Radical Axis

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 x2 + y2 + 2gx + 2fy + c=0 and
S' º x2 + y2 + 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.

Example 1

Prove that the circle x2 + y2 – 6x – 4y + 9 = 0 bisects the circumference of the circle x2 + y2 – 8x – 6y + 23 = 0.

Solution:

The given circles are

S1 º x2 + y2 – 6x – 4y + 9 = 0 ………..(1) .

and S2 º x2 + y2 – 8x – 6y + 23 = 0. …….….(2).

Equation of the common chord of circles (1) and (2) which is also the radical axis of the circles S1 and S­2 is

S1 – S2 = 0 or, 2x + 2y – 14 = 0

or, x + y – 7 = 0 ………..(3).

Centre of the circle S2 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).

Example 2

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

Solution:

Let us take the equation of the two circles as

x2 + y2 + 2l1x + a = 0 ….(1) .

x2 + y2 + 2l2x + 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 x2 + y2 + 2gx + 2fy + c = 0 ….(3) .

The circle (1) and (3) cut orthogonally 2l1g = a + c ….(4) .

The circles (2) and (3) cut orthogonally

2l2 g = a + c …(5)

From (4) and (5), 2g(l1 – l2) = 0 but l1 > l2

\ g = 0

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

The common chord of (1) and (2) has the equation S1 – S2 = 0

i.e. (x2 + y2 + 2l1x + a) – (x2 + y2 + 2l2x + a) = 0 .

or 2(l1 – l2) x = 0 \ x = 0

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

 
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