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Let A B be chord of contact of the point...

Let `A B` be chord of contact of the point `(5,-5)` w.r.t the circle `x^2+y^2=5` . Then find the locus of the orthocentre of the triangle `P A B` , where `P` is any point moving on the circle.

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Equation of chord of contact AB of circle `x^(2)+y^(2)=5` w.r.t. point (5,-5) is
`5x-5y=5` or `x-y=1`.
Solving with the circle we get
`x^(2)+(x-1)^(2)=5`
`implies x^(2)-x-2=0`
`implies (x-2)(x+1)=0`
`implies x= -1,2`
`implies y= -2,1`
So, chord of contact meets the circle at A(-1,-2) and B(2,1).
Let point P on the circle be `(sqrt(5) cos theta, sqrt(5) sin theta)`.
Now, as circumcentre of the traingle PAB is origin.
For orthocentre (h,k), we have
`h= -1+2+sqrt(5) cos theta ` and `k= -2+1+sqrt(5) sin theta`
or `h-1= sqrt(5) cos theta ` and `k+1 = sqrt(5) sin theta`
Squaring and adding , we get `(h-1)^(2)+(k+1)^(2)=5`.
Hence, required locus is `(x-1)^(2)+(y+1)^(2)=5`
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