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The equation of the cirele which passes ...

The equation of the cirele which passes through the point (1, 1) and touches the circle `x^2+y^2+4x-6y-3=0` at the point (2, 3) on it is

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Given circle is
`x^(2)+y^(2)+4x-6y-3=0` (1)
Required circle touches above circle at point P(2,3).
So, it touches the tangent at point P to the given circle.
Equation of tangent at point P(2,3) to the given circle is
`2x+3y+2(x+2)-3(y+3)-3=0`
or `x-2=0`
Equatiion of family of circle touching line `x-2=0` at (2,3) is given by
`[(x-2)^(2)+(y-3)^(2)]+lambda(x-2)=0,lambda in R` (2)
It if passes through (1,1), then
`1+4+lambda(-1)=0` or `lambda=5`
Putting the value of `lambda` in equation (2), we get
`(x-2)^(2)+(y-3)^(2)+5(x-3)=0`
or `x^(2)+y^(2)+x-6y+3=0`
This is the required circle.
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