The equation of the circle having the chord x | vedclass

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(C) The equation of a family of circles passing through the intersection of the circle $S \equiv 2x^2 + 2y^2 - 2x - 6y - 25 = 0$ and the line $L \equi

Vedclass · Accepted Answer

$2x^2 + 2y^2 - 6x - 2y - 21 = 0$

Vedclass · Answer

(C) The equation of a family of circles passing through the intersection of the circle $S \equiv 2x^2 + 2y^2 - 2x - 6y - 25 = 0$ and the line $L \equiv x - y - 1 = 0$ is given by $S + \lambda L = 0$.$(2x^2 + 2y^2 - 2x - 6y - 25) + \lambda(x - y - 1) = 0$$2x^2 + 2y^2 + (\lambda - 2)x - (\lambda + 6)y - (25 + \lambda) = 0$The center of this circle is $\left( -\frac{\lambda - 2}{4}, \frac{\lambda + 6}{4} \right)$.Since the line $x - y - 1 = 0$ is the diameter,the center must lie on this line:$-\frac{\lambda - 2}{4} - \frac{\lambda + 6}{4} - 1 = 0$$-\lambda + 2 - \lambda - 6 - 4 = 0$$-2\lambda - 8 = 0 \implies \lambda = -4$.Substituting $\lambda = -4$ into the family equation:$2x^2 + 2y^2 + (-4 - 2)x - (-4 + 6)y - (25 - 4) = 0$$2x^2 + 2y^2 - 6x - 2y - 21 = 0$.

The equation of the circle having the chord $x - y - 1 = 0$ of the circle $2x^2 + 2y^2 - 2x - 6y - 25 = 0$ as its diameter is:

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