Let a circle pass through the origin and its centre be the point of intersection of the tw…

Question

Let a circle pass through the origin and its centre be the point of intersection of the two mutually perpendicular lines $x+(k-1)y+3=0$ and $2x+k^2y-4=0$. If the line $x-y+2=0$ intersects the circle at points $A$ and $B$, then $(AB)^2$ is equal to:

Accepted Answer

Perpendicularity: $\left(\dfrac{1}{1-k}\right)\left(\dfrac{2}{k^2}\right)=1 \Rightarrow 2=k^2-k^3\Rightarrow k^3-k^2+2=0\Rightarrow k=-1$.

The lines become $2(x-2y+3)=0$ and $2x+y-4=0$; solving gives centre $(1,2)$, and since the circle passes through the origin, $r=\sqrt5$: $(x-1)^2+(y-2)^2=5$.

Distance from $(1,2)$ to $x-y+2=0$: $p=\dfrac{|1-2+2|}{\sqrt2}=\dfrac{1}{\sqrt2}$.

$$AB=2\sqrt{r^2-p^2}=2\sqrt{5-\tfrac12}=2\sqrt{\tfrac92}=\frac{6}{\sqrt2},\qquad AB^2=18.$$

Solutions · 1