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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.

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The correct Answer is:
(6,-4)
Any circle which touches the line `x+y-2=0` at (1,1) will be of the form
`(x-1)^(2)+(y-1)^(2)+lambda(x+y-2)=0`
or `x^(2)+y^(2)+(lambda-2)x+(lambda-2)y+2-2lambda=0`
The common chord of this circle and `x^(2)+y^(2)+4x+5y-6=0` will be
`(lambda-6)x+(lambda-7)y+8-2lambda=0`
or `(-6-7y+8)+lambda(x+y-2)=0`
which is a family of lines, each member of which will be passing through a fixed , which is the point of intersection of the lines `-6x-7y+8=0` and `x+y-2=0, i.e., (6,-4)`.
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