The line $x+y=k$ meets the curve $x^2+y^2-2 x-4 y+2=0$ at two points $\mathrm{A}$ and $\mathrm{B}$. If $O$…
The line $x+y=k$ meets the curve $x^2+y^2-2 x-4 y+2=0$ at two points $\mathrm{A}$ and $\mathrm{B}$. If $O$ is the origin and $\angle \mathrm{AOB}=90^{\circ}$, then the value of $k(k>1)$ is
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Solution
Eqn. of circle is
$x^2+y^2-2 x-4 y+2=0$ ...(i)
Homogenising eqn. (i), we get
$x^2+y^2-2 x \times 1-4 y \times 1+2 \times(1)^2=0$
$\begin{aligned} \Rightarrow x^2+y^2-2 x\left(\frac{x+y}{k}\right)-4 y\left(\frac{x+y}{k}\right)+2\left(\frac{x+y}{k}\right)^2=0 & \\ & \left\{\because x+y=k \Rightarrow \frac{x+y}{k}=1\right\}\end{aligned}$
$\begin{array}{r}\Rightarrow k^2 x^2+k^2 y^2-2 k x^2-2 k x y-4 k x y-4 k y^2+2 x^2 \\ +2 y^2+4 x y=0\end{array}$
$\begin{aligned} & \Rightarrow x^2\left(k^2-2 k+2\right)+(4-6 k) x y+\left(k^2-4 k+2\right) y^2=0 \\ & \text { Given : } \angle A O B=90^{\circ} \\ & \Rightarrow \text { Coefficient } x^2+\text { Coefficient of } y^2=0 \\ & \Rightarrow k^2-2 k+2+k^2-4 k+2=0\end{aligned}$
$\begin{aligned} & \Rightarrow 2 k^2-6 k+4=0 \\ & \Rightarrow k^2-3 k+2=0 \\ & \Rightarrow(k-2)(k-1)=0 \\ & \Rightarrow k=2,1 \\ & \text { Given, } k>1 \Rightarrow k=2 .\end{aligned}$