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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

Verified by Experts

The correct Answer is:
`x^(2)+y^(2)+18x-2y+32=0`
The parametric form of OP is
`(x-0)/(cos45^(@))=(y-0)/(sin45^(@))`
Since, `OP = 4sqrt2`
So, the coordinates of P are given by
`(x-0)/(cos45^(@))=(y-0)/(sin45^(@))=-4sqrt2`
So, P(-4, - 4)
Let C (h,k) be the centre of circle and r be its radius Now, `CPbotOP`

`rArr (k+4)/(h+4)*(1)=-1`
`rArr K+4=-h-4`
`rArr h+k=-8" "...(i)`
`Also, CP^(2)=(h+4)^(2)+(k+4)^(2)`
`rArr (h+4)^(2)+(k+4)^(2)=r^(2)" "...(ii)`
In `DeltaACM`, we have
`AC^(2)=(3sqrt2)^(2)+((h+k)/(sqrt2))^(2)`
`rArr r^(2)=18+32`
`rArrr = 5 sqrt2" "...(iii)`
Also, CP = r
`rArr h-k=pm 10" "...(iv)`
From, the equation of the circles are
`(x+9)^(2)+(y-1)^(2)=(5sqrt2)^(2)`
or `(x-1)^(2)+(y+9)^(2)=(5sqrt2)^(2)`
`rArr x^(2) +y^(2)+ 18x - 2y + 32 =0 `
`or x^(2)+ y^(2)-2x +18y + 32 = 0`
Clearly, (-10, 2) lies interior of
`x^(2)+y^(2)+18x-2y+32=0`
Hence, the required equation of circle, is
`x^(2)+y^(2)+18x-2y+32=0`
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