Find the equation of the circle whose diamete | vedclass

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(C) Let the two circles be $S_1: x^2 + y^2 - 8x + y - 15 = 0$ and $S_2: x^2 + y^2 - 4x + 4y - 42 = 0$.The equation of the common chord is given by $S_

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$x^2 + y^2 - 12x - 2y + 12 = 0$

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(C) Let the two circles be $S_1: x^2 + y^2 - 8x + y - 15 = 0$ and $S_2: x^2 + y^2 - 4x + 4y - 42 = 0$.The equation of the common chord is given by $S_1 - S_2 = 0$.$(x^2 + y^2 - 8x + y - 15) - (x^2 + y^2 - 4x + 4y - 42) = 0$.$-4x - 3y + 27 = 0$ or $4x + 3y - 27 = 0$.The family of circles passing through the intersection of $S_1$ and $S_2$ is $S_1 + \lambda(S_1 - S_2) = 0$.$(x^2 + y^2 - 8x + y - 15) + \lambda(4x + 3y - 27) = 0$.$x^2 + y^2 + x(4\lambda - 8) + y(3\lambda + 1) - 27\lambda - 15 = 0$.The center of this circle is $(- (2\lambda - 4), - \frac{3\lambda + 1}{2})$.Since the common chord is the diameter,the center must lie on the line $4x + 3y - 27 = 0$.$4(-2\lambda + 4) + 3(-\frac{3\lambda + 1}{2}) - 27 = 0$.$-8\lambda + 16 - \frac{9\lambda + 3}{2} - 27 = 0$.$-16\lambda + 32 - 9\lambda - 3 - 54 = 0$.$-25\lambda - 25 = 0 \implies \lambda = -1$.Substituting $\lambda = -1$ into the equation:$x^2 + y^2 + x(-4 - 8) + y(-3 + 1) + 27 - 15 = 0$.$x^2 + y^2 - 12x - 2y + 12 = 0$.

Find the equation of the circle whose diameter is the common chord of the circles $x^2 + y^2 - 8x + y - 15 = 0$ and $x^2 + y^2 - 4x + 4y - 42 = 0$.

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