The equation of the circle having the chord x | vedclass

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(A) The equation of a family of circles passing through the intersection of a circle $S = 0$ and a line $L = 0$ is given by $S + \lambda L = 0$. Here,

Vedclass · Accepted Answer

$x^2 + y^2 - a^2 - 2p(x \cos \alpha + y \sin \alpha - p) = 0$

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(A) The equation of a family of circles passing through the intersection of a circle $S = 0$ and a line $L = 0$ is given by $S + \lambda L = 0$. Here,$S = x^2 + y^2 - a^2 = 0$ and $L = x \cos \alpha + y \sin \alpha - p = 0$. So,the equation of the circle is $(x^2 + y^2 - a^2) + \lambda(x \cos \alpha + y \sin \alpha - p) = 0$. Since this circle has the line $x \cos \alpha + y \sin \alpha - p = 0$ as its diameter,the center of this circle must lie on the line. The center of the circle $x^2 + y^2 + (\lambda \cos \alpha)x + (\lambda \sin \alpha)y - (a^2 + \lambda p) = 0$ is $(-\frac{\lambda \cos \alpha}{2}, -\frac{\lambda \sin \alpha}{2})$. Substituting this center into the line equation: $(-\frac{\lambda \cos \alpha}{2}) \cos \alpha + (-\frac{\lambda \sin \alpha}{2}) \sin \alpha - p = 0$ $-\frac{\lambda}{2} (\cos^2 \alpha + \sin^2 \alpha) = p$ $-\frac{\lambda}{2} = p \implies \lambda = -2p$. Substituting $\lambda = -2p$ into the family equation,we get: $x^2 + y^2 - a^2 - 2p(x \cos \alpha + y \sin \alpha - p) = 0$.

The equation of the circle having the chord $x \cos \alpha + y \sin \alpha = p$ of the circle $x^2 + y^2 = a^2$ as its diameter is:

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