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A circle touches the line y = x at point...

A circle touches the line `y = x` at point P such that `OP = 4 sqrt2`, Circle contains (-10,2) in its interior & length of its chord on the line `x+y=0` is ` 6sqrt2`. Determine the equation of the circle

Text Solution

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let the point on circle which lies on the line x=y may be `(4,4) and (-4,-4)`
As the point (-10,2) lies inside the circle so, the required point on circle be`(-4,-4)`
Line perpendicular to y=x and passes through (-4,-4) be
`L1:x+y=-8`
Distance between the given chord and the line L1 is=`|(-8-0)/sqrt(2)|=8/sqrt2=4sqrt2`
By applying pythagoras theorem in `/_ CBD`, `R^2=(3sqrt2)^2+(4sqrt2)^2` `R=5sqrt2`
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