The locus of the midpoints of the chords of t | vedclass

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(C) The given circle is $x^2 + y^2 - ax - by = 0$. Its center is $C = \left( \frac{a}{2}, \frac{b}{2} \right)$ and its radius is $R = \sqrt{\left( \fr

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$x^2 + y^2 - ax - by + \frac{a^2 + b^2}{8} = 0$

Vedclass · Answer

(C) The given circle is $x^2 + y^2 - ax - by = 0$. Its center is $C = \left( \frac{a}{2}, \frac{b}{2} \right)$ and its radius is $R = \sqrt{\left( \frac{a}{2} \right)^2 + \left( \frac{b}{2} \right)^2} = \frac{\sqrt{a^2 + b^2}}{2}$.Let $(h, k)$ be the midpoint of a chord. The distance from the center $C$ to the chord is $d = \sqrt{(h - a/2)^2 + (k - b/2)^2}$.Since the chord subtends a right angle at the center,the triangle formed by the center and the endpoints of the chord is an isosceles right-angled triangle.Thus,the distance $d$ from the center to the chord is $d = R \sin(45^\circ) = \frac{R}{\sqrt{2}}$.Squaring both sides,$d^2 = \frac{R^2}{2}$.Substituting the values: $\left( h - \frac{a}{2} \right)^2 + \left( k - \frac{b}{2} \right)^2 = \frac{1}{2} \left( \frac{a^2 + b^2}{4} \right) = \frac{a^2 + b^2}{8}$.Expanding this,$h^2 - ah + \frac{a^2}{4} + k^2 - bk + \frac{b^2}{4} = \frac{a^2 + b^2}{8}$.$h^2 + k^2 - ah - bk + \frac{a^2 + b^2}{4} - \frac{a^2 + b^2}{8} = 0$.$h^2 + k^2 - ah - bk + \frac{a^2 + b^2}{8} = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 - ax - by + \frac{a^2 + b^2}{8} = 0$.

The locus of the midpoints of the chords of the circle $x^2 + y^2 - ax - by = 0$ which subtend a right angle at $\left( \frac{a}{2}, \frac{b}{2} \right)$ is:

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