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7. Prove that the locus of the poles of chords of the circle $x^2 + y^2 = a^2$ which subtend a right angle at the fixed point $(h, k)$ is the circle
$$ (h^2+k^2-a^2)(x^2+y^2)-2a^2ky+2a^4 = 0.$$

Solution:

The polar of $(X, Y)$ is $xX + yY = a^2$. Transferring to $(h,k)$ as origin, this becomes
$$xX + yY = a^2-hX-kY .......................... (i),$$
while the equation to the circle is now
$$x^2+y^2+2hx+2ky+h^2+k^2-a^2 = 0 .......................... (ii).$$

Forming the equation to the lines joining the origin to the intersections of $(i)$ and $(ii)$ by the ordinary rule, and writing down the condition that these are perpendicular, we obtain the required locus of $(X, Y)$.

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