CircleHardBloom L3
Question
The locus of the mid points of the chords of the circle $x^{2} + y^{2} - ax - by = 0$ which subtend a right angle at ( $a/2,b/2$ ) is:
Options
A.$ax + by = 0$
B.$ax + by = a^{2} + b^{2}$
C.$x^{2} + y^{2} - ax - by + \frac{a^{2} + b^{2}}{8} = 0$
D.$x^{2} + y^{2} - ax - by - \frac{a^{2} + b^{2}}{8} = 0$
Solution
[Key Concept: Locus of midpoints of chords subtending a right angle at the centre: use $T = S_{1}$ with the angle condition.]
$OA^{2} = AC^{2} + OC^{2}$
$$\begin{matrix} & \frac{a^{2}}{4} + \frac{b^{2}}{4} = 2\left\lbrack \left( h - \frac{a}{2} \right)^{2} + \left( k - \frac{b}{2} \right)^{2} \right\rbrack \\ \Rightarrow & h^{2} + k^{2} - ah - bk + \frac{a^{2}}{8} + \frac{b^{2}}{8} = 0 \end{matrix}$$
Required Locus is
$$x^{2} + y^{2} - ax - by + \frac{a^{2} + b^{2}}{8} = 0$$
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