The equation of the circle on the common chord of the circles (x – a)2 + y2 = a2 and x2 + (y + b)2 = b2 as diameter is
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x2 + y2 = 2ab(bx + ay)
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x2 + y2 = bx + ay
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(a2 + b2)(x2 + y2) = 2ab(bx – ay)
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(a2 + b2) (x2 + y2) = 2(bx + ay)
The common chord of (x-a)^2 + y^2 = a^2 and x^2 + (y+b)^2 = b^2 is found by subtracting the equations: x^2 - 2ax + a^2 + y^2 = a^2 and x^2 + y^2 + 2by + b^2 = b^2. x^2 + y^2 - 2ax = 0 and x^2 + y^2 + 2by = 0. Subtracting gives 2ax + 2by = 0, or ax + by = 0. The circle with this chord as diameter is (x-x1)(x-x2) + (y-y1)(y-y2) = 0. The derived equation matches option C.
The first circle is x^2 - 2ax + y^2 = 0 and the second is x^2 + y^2 + 2by = 0. Subtracting them yields the common chord, 2by + 2ax = a^2 - b^2. The required circle passing through this chord's endpoints (as diameter) is given by S1 + lambda(S2 - S1) = 0. Solving this family of circles by expanding and setting the coefficient of x equal to negative the coefficient of y yields the equation (a^2 + b^2)(x^2 + y^2) = 2ab(bx - ay).