The condition that the lines joining the origin to the points of intersection of the line…
The condition that the lines joining the origin to the points of intersection of the line $\frac{x}{a}+\frac{y}{b}=2$ and the circle $(x-a)^2+(y-b)^2=r^2$ are at right angles is
$a^2+b^2=r^2$
$a^2-b^2=r^2$
$a^2-b^2+r^2=0$
$\mathrm{a}^2+\mathrm{b}^2+\mathrm{r}^2=0$
Solution
$\because \frac{x}{a}+\frac{y}{b}=2 \Rightarrow \frac{x}{2 a}+\frac{y}{2 b}=1$
$
\begin{aligned}
& \&(x-a)^2+(y-b)^2=r^2 \\
& \Rightarrow x^2+y^2-2 a x-2 b y+a^2+b^2-r^2=0
\end{aligned}
$
Homogenizing eqn. (ii), we get
$
\begin{aligned}
& \Rightarrow x^2+y^2-2(a x+b y) 1+\left(a^2+b^2-r^2\right) 1^2=0 \\
& \Rightarrow x^2+y^2-2(a x+b y)\left(\frac{x}{2 a}+\frac{y}{2 b}\right)+\left(a^2+b^2-r^2\right) \\
& \left(\frac{x}{2 a}+\frac{y}{2 b}\right)^2=0 \\
& \Rightarrow 4 a^2 b^2\left(x^2+y^2\right)-4 a b(a x+b y)(b x+a y)+ \\
& \left(a^2+b^2-r^2\right)(b x+a y)^2=0 \\
& \Rightarrow 4 a^2 b^2\left(x^2+y^2\right)-4 a b\left(a b x^2+a b y^2+\left(a^2+b^2\right) x y\right)+ \\
& \left(a^2+b^2-r^2\right)(b x+a y)^2=0 \\
& \Rightarrow 4 a^2 b^2\left(x^2+y^2\right)-4 a^2 b^2\left(x^2+y^2\right)-4 a b \\
& \left(a^2+b^2\right) x y+\left(a^2+b^2-r^2\right)\left(b^2 x^2+a^2 y^2\right)+ \\
& \left.\left(a^2+b^2-r^2\right) 2 a b x y\right)=0
\end{aligned}
$
This is the equation of the pair of straight lines joining the origin to the points of intersection of the given lines \& the curve. They will be at right angle if the sum of coefficients of $x^2$ and $y^2$ is zero.
Imposing above condition, we get
$
\begin{aligned}
& \left(a^2+b^2-r^2\right)\left(a^2+b^2\right)=0 \\
& \Rightarrow a^2+b^2-r^2=0 \\
& \Rightarrow a^2+b^2=r^2
\end{aligned}
$