# 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 ± k B) c^{2}=h^{2}+k^{2} C) c^{2}=(h+k)^{2} D) 4c^{2}=h^{2}+k^{2}

## 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}$

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## Answer

The correct option is **B**

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

Given the circle equation: $$ (x - h)^2 + (y - k)^2 = c^2 $$ and the line equation: $$ kx + hy = 2hk $$ we can rewrite the line equation in the intercept form: $$ \frac{x}{2h} + \frac{y}{2k} = 1 $$

Substitute $x$ and $y$ from the line into the circle’s equation and make the equation homogeneous: $$ \left(x^2 + y^2\right) - 2\left(hx + ky\right)\left(\frac{x}{2h} + \frac{y}{2k}\right) + \left(h^2 + k^2 - c^2\right)\left(\frac{x}{2h} + \frac{y}{2k}\right)^2 = 0 $$

For the resulting lines to be perpendicular:

The coefficient sum of $x^2$ and $y^2$, which are associated with the slopes of the lines, must be $0$.

From simplifying the equation, we get: $$ \left[1 - 1 + \frac{h^2 + k^2 - c^2}{4h^2}\right] + \left[1 - 1 + \frac{h^2 + k^2 - c^2}{4k^2}\right] = 0 $$ or $$ (h^2 + k^2 - c^2)\left(\frac{h^2 + k^2}{4h^2 k^2}\right) = 0 $$

Thus, to satisfy the above: $$ h^2 + k^2 = c^2 $$

This confirms that the correct answer is **Option B: $c^2 = h^2 + k^2$**.

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