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(A) Let the midpoint of the chord $AB$ of the circle $x^2 + y^2 = a^2$ be $P(h, k)$.The equation of the chord with midpoint $(h, k)$ is given by $T =

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$2(x^2 + y^2) - a^2 = 0$

Vedclass · Answer

(A) Let the midpoint of the chord $AB$ of the circle $x^2 + y^2 = a^2$ be $P(h, k)$.The equation of the chord with midpoint $(h, k)$ is given by $T = S_1$,which is $hx + ky = h^2 + k^2$.The combined equation of the lines $OA$ and $OB$ is obtained by homogenizing the circle equation $x^2 + y^2 = a^2$ using the chord equation:$x^2 + y^2 = a^2 \left( \frac{hx + ky}{h^2 + k^2} \right)^2$$(h^2 + k^2)(x^2 + y^2) = a^2(hx + ky)^2$$(h^2 + k^2)(x^2 + y^2) - a^2(h^2x^2 + k^2y^2 + 2hkxy) = 0$$(h^2 + k^2 - a^2h^2)x^2 + (h^2 + k^2 - a^2k^2)y^2 - 2a^2hkxy = 0$Since the chord subtends a right angle at the center,the sum of the coefficients of $x^2$ and $y^2$ must be zero:$(h^2 + k^2 - a^2h^2) + (h^2 + k^2 - a^2k^2) = 0$$2(h^2 + k^2) - a^2(h^2 + k^2) = 0$$(h^2 + k^2)(2 - a^2) = 0$Wait,re-evaluating: The distance from the center $(0,0)$ to the chord $hx + ky = h^2 + k^2$ is $d = \frac{|h^2 + k^2|}{\sqrt{h^2 + k^2}} = \sqrt{h^2 + k^2}$.In a right-angled triangle formed by the center and the chord,the distance $d$ from the center to the chord is $a \cos(45^\circ) = \frac{a}{\sqrt{2}}$.So,$\sqrt{h^2 + k^2} = \frac{a}{\sqrt{2}}$.Squaring both sides,$h^2 + k^2 = \frac{a^2}{2}$,which gives $2(h^2 + k^2) - a^2 = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $2(x^2 + y^2) - a^2 = 0$.

Find the locus of the midpoint of a chord of the circle $x^2 + y^2 = a^2$ which subtends a right angle at the center.

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