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 $2x + k^{2}y - 4 = 0$. If the line $x - y + 2 = 0$ intersects the circle at the points A and B , then $(AB)^{2}$ is equal to :
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Solution
Sol. [Key Concept: Centre $=$ intersection of the two perpendicular lines; chord $AB$ from the perpendicular-distance formula.]
$${x + (k - 1)y + 3 = 0 }{2x + k^{2}y - 4 = 0 }{\left( \frac{1}{1 - k} \right)\left( \frac{2}{k^{2}} \right) = 1 }{2 = k^{2} - k^{3} }{k^{3} - k^{2} + 2 = 0 }{k = - 1 }$$Solving :
$${2(x - 2y + 3) = 0 }{2x + y - 4 = 0 }{- 5y + 10 = 0 }{y = 2 }$$
$$x = 1 $$Centre $(1,2)$
$$r = \sqrt{5}$$
So circle is $(x - 1)^{2} + (y - 2)^{2} = 5$
Chord $x - y + 2 = 0$
$$\begin{matrix} & p = \frac{1}{\sqrt{2}} \\ & \mathcal{l} = \sqrt{5 - \frac{1}{2}} = \frac{3}{\sqrt{2}} \\ & AB = \frac{6}{\sqrt{2}}\ \therefore{AB}^{2} = 18 \end{matrix}$$
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