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

Question

Let a circle pass through the origin and its centre be the point of intersection of 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 the points A and B, then...

Accepted Answer

$\begin{gathered} x+(k-1)y+3=0 \ 2x+{k}^{2}y-4=0 \ (\frac{1}{1-k})(\frac{2}{{k}^{2}})=1 \ 2={k}^{2}-{k}^{3} \ {k}^{3}-{k}^{2}+2=0 \ k=-1 \end{gathered}$

Solving :

$\begin{gathered} 2(x-2y+3)=0 \ 2x+y-4=0 \ -5y+10=0 \ y=2 \end{gathered}$

$x=1$

Centre $(1,2)$

$r=\sqrt{5}$

So circle is $(x-1{)}^{2}+(y-2{)}^{2}=5$

Chord $x-y+2=0$

$\begin{gathered} p=\frac{1}{\sqrt{2}} \ l=\sqrt{5-\frac{1}{2}}=\frac{3}{\sqrt{2}} \ AB=\frac{6}{\sqrt{2}}\, 	herefore {AB}^{2}=18 \end{gathered}$

Solutions · 1