Prove that the straight lines joining the origin to the points of intersection of the straight line
kx + hy = 2hk
with the curve $(x - h)^{2} + (y - k)^{2} = c^{2}$
are at right angles if $h^{2} + k^{2} = c^{2} .$

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Solution

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

The equation of pair straight line joining the origin is find by method of homogenisation. We make the coefficient of constant term of straight line $1$ and put it the equation of the curve so that degree of each term of the curve become $2$

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

Lines are prerpendicular if $a + b = 0$

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

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Prove that the straight lines joining the origin to the points of intersection of the straight linekx + hy = 2hk with the curve (x−h)2+(y−k)2=c2are at right angles if h2+k2=c2.

Prove that the straight lines joining the origin to the points of intersection of the straight line
kx + hy = 2hk
with the curve $(x - h)^{2} + (y - k)^{2} = c^{2}$
are at right angles if $h^{2} + k^{2} = c^{2} .$