A circle touches the line `y = x` at the point `(2, 2)` and has it centre on y-axis, then square of its radius is

Answer

8

Explanation

A circle touches …………
Equation of a circle touching the line `y = x at (2, 2)` is
`(x-2)^(2)+(y-2)^(2)+lambda(x-y) = 0`
`i.e. x^(2)y^(2)-(4-lambda)x- (4+lambda)y +8 =0`
its centre `((4-lambda)/(2), (4+lambda)/(2))` lies on y-axis
`:. (4-lambda)/(2) = 0` i.e. `lambda = 4`
`:.` the circle is `x^(2)+y^(2)8y+8=0`
`:.` radius `= sqrt(16-8) = sqrt(8)`