Show that the equation of the circle which has for its diameter, the chord cut off by the straight line $a x + b y + c = 0$ by the circle
$(a^{2} + b^{2}) (x^{2} + y^{2}) = 2 c^{2}$ is
$(a^{2} + b^{2}) (x^{2} + y^{2}) + 2 c (a x + b y) = 0$

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Solution

Apply $S + λ = 0$
$(x^{2} + y^{2} - \frac{2 c^{2}}{a^{2} + b^{2}}) + λ (a x + b y + c) = 0$
Its centre $(- \frac{λ a}{2}, - \frac{λ b}{2})$ lies on diameter $a x + b y + c = 0 ∴ λ \frac{2 c}{a^{2} + b^{2}}$ etc

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

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