Problem: The straight lines joining the origin to the points of intersection of the straight line hx + ky = 2hk and the curve (x – k)^{2} + (y – h)^{2} = c^{2} are at right angles then

(A) h^{2} + k^{2} + c^{2}= 0

(B) h^{2} – k^{2} – c^{2} = 0

(C) h^{2}+ k^{2} – c^{2} = 0

(D) none of these

Ans: (C)

Sol: ). Making the equation of the curve homogeneous with the help of the line, we get

$\large x^2 + y^2 -2(kx+hy)(\frac{hx+ky}{2hk}) + (h^2 + k^2 -c^2)(\frac{hx+ky}{2hk})^2 = 0 $

4h^{2}k^{2}x^{2}+4h^{2}k^{2}y^{2}–4hk^{2}x(hx+ky)–4h^{2}ky(hx+ky)+(h^{2}+k^{2}-c^{2})(h^{2}x^{2}+k^{2}y^{2}+ 2hkxy)=0

This is the equation of the pair of lines joining the origin to the points of intersection of the given line and the curve.

They will be at right angles if coefficient of x^{2} + coefficient of y^{2} = 0

(h^{2} + k^{2}) (h^{2} + k^{2} – c^{2}) = 0 since [h^{2} + k^{2} ≠ 0]

⇒ h^{2} + k^{2} = c^{2}.