Question

The lines joining the points of intersection of the curve $(x - h)^{2} + (y - k)^{2} - c^{2} = 0$ and the line $kx + hy = 2hk$ to the origin are perpendicular, then

• A. $c = h \pm k$
• B. $c^{2} = h^{2} + k^{2}$
• C. $c^{2} = (h + k)^{2}$
• D. $4c^{2} = h^{2} + k^{2}$

Answer 1

Answer: (B)

$c^{2} = h^{2} + k^{2}$

Answer 2

The line is $\frac{x}{2h} + \frac{y}{2k} = 1$and circle is,

$x^{2} + y^{2} - 2(hx + ky) + (h^{2} + k^{2} - c^{2}) = 0$

Making it homogeneous, we get

$\Rightarrow (x^{2} + y^{2}) - 2(hx + ky)\left( \frac{x}{2h} + \frac{y}{2k} \right) + (h^{2} + k^{2} - c^{2})\left( \frac{x}{2h} + \frac{y}{2k} \right)^{2} = 0$

If these lines be perpendicular, then $A + B = 0$

$\left\lbrack 1 - 1 + \frac{(h^{2} + k^{2} - c^{2})}{4h^{2}} \right\rbrack + \left\lbrack 1 - 1 + \frac{(h^{2} + k^{2} - c^{2})}{4k^{2}} \right\rbrack = 0$

or $(h^{2} + k^{2} - c^{2})\ \left( \frac{h^{2} + k^{2}}{4h^{2}k^{2}} \right) = 0$

$\therefore h^{2} + k^{2} = c^{2}$.