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

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Solution

The correct option is B

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

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

Making it homogeneous, we get

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

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

$[1 - 1 + \frac{(h^{2} + k^{2} - c^{2})}{4 h^{2}}] + [1 - 1 + \frac{(h^{2} + k^{2} - c^{2})}{4 k^{2}}] = 0$

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

$∴ h^{2} + k^{2} = c^{2}$ .

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The lines joining the points of intersection of the curve (x−h)2+(y−k)2−c2=0 and the line kx + hy = 2hk to the origin are perpendicular, then

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