The lines joining the points of intersection | vedclass

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(B) The equation of the circle is $x^2 + y^2 - 2hx - 2ky + h^2 + k^2 - c^2 = 0$. The equation of the line is $\frac{kx + hy}{2hk} = 1$,which is $\frac

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

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(B) The equation of the circle is $x^2 + y^2 - 2hx - 2ky + h^2 + k^2 - c^2 = 0$. The equation of the line is $\frac{kx + hy}{2hk} = 1$,which is $\frac{x}{2h} + \frac{y}{2k} = 1$. Making the circle equation homogeneous using the line equation: $(x^2 + y^2) - 2(hx + ky)\left(\frac{x}{2h} + \frac{y}{2k}ight) + (h^2 + k^2 - c^2)\left(\frac{x}{2h} + \frac{y}{2k}ight)^2 = 0$. Expanding this,the coefficient of $x^2$ is $A = 1 - 1 + \frac{h^2 + k^2 - c^2}{4h^2} = \frac{h^2 + k^2 - c^2}{4h^2}$. The coefficient of $y^2$ is $B = 1 - 1 + \frac{h^2 + k^2 - c^2}{4k^2} = \frac{h^2 + k^2 - c^2}{4k^2}$. For the lines to be perpendicular,$A + B = 0$. $(h^2 + k^2 - c^2) \left(\frac{1}{4h^2} + \frac{1}{4k^2}ight) = 0$. Since $\frac{1}{4h^2} + \frac{1}{4k^2} 
eq 0$,we must have $h^2 + k^2 - c^2 = 0$,or $c^2 = h^2 + k^2$.

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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