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A variable circle which always touches t...

A variable circle which always touches the line `x+y-2=0` at (1, 1) cuts the circle `x^2+y^2+4x+5y-6=0` . Prove that all the common chords of intersection pass through a fixed point. Find that points.

A

PQ can never be parallel to the given line `x +y - 2 = 0`

B

PQ can never be perpendicular to the given line `x +y - 2 = 0`

C

PQ always passes through `(6,-4)`

D

PQ always passes through `(-6,4)`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
Family of circles touching `x +y -2 =0` at (1,1) is `(x-1)^(2)+ (y-1)^(2) + lambda (x+y-2) = 0`
`rArr x^(2) +y^(2) + (lambda-2) x +(lambda-2) y + 2 -2 lambda = 0` (1)
Given circle is:
`x^(2) +y^(2) +4x +5y - 6 = 0` (2)
Equation of common chord PQ is `S- S' = 0`.
`rArr (lambda -6)x + (lambda -7)y +8 - 2lambda = 0` (3)
(a) PQ || line `x +y - 2 = 0`
`rArr (6-lambda)/(lambda-7) =- 1 rArr 6 = 7`, which is impossible
(b) `PQ _|_` line: `x +y -2 = 0`
`rArr (6-lambda)/(lambda-7) =1 rArr lambda = (13)/(2)`, which is possible
But when `lambda = (13)/(2)`, we can see that the circles (1) and (2) are not intersecting each other and their radical axis is perpendicular to the given line `x +y -2 = 0`.
Eq. (3) can be written as
`-6x - 7y +8 + lambda (x+y-2) =0`
which is in the form `L_(1) + lambda L_(2) =0`
Solving `L_(1)` and `L_(2)`we get the point of intersection (6,-4).
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