The equation of a circle is x^2 + y^2 = a^2 a | vedclass

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(A) The equation of any circle passing through the intersection of the circle $x^2 + y^2 - a^2 = 0$ and the line $x \cos \alpha + y \sin \alpha - p =

Vedclass · Accepted Answer

$x^2 + y^2 - 2px \cos \alpha - 2py \sin \alpha + 2p^2 - a^2 = 0$

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(A) The equation of any circle passing through the intersection of the circle $x^2 + y^2 - a^2 = 0$ and the line $x \cos \alpha + y \sin \alpha - p = 0$ is given by the family of circles equation:$x^2 + y^2 - a^2 + \lambda(x \cos \alpha + y \sin \alpha - p) = 0$The center of this circle is $\left(-\frac{\lambda \cos \alpha}{2}, -\frac{\lambda \sin \alpha}{2}\right)$.Since the line $x \cos \alpha + y \sin \alpha = p$ is a diameter,the center must lie on this line:$\left(-\frac{\lambda \cos \alpha}{2}\right) \cos \alpha + \left(-\frac{\lambda \sin \alpha}{2}\right) \sin \alpha = p$$-\frac{\lambda}{2} (\cos^2 \alpha + \sin^2 \alpha) = p$$-\frac{\lambda}{2} = p \implies \lambda = -2p$Substituting $\lambda = -2p$ into the family equation:$x^2 + y^2 - a^2 - 2p(x \cos \alpha + y \sin \alpha - p) = 0$$x^2 + y^2 - 2px \cos \alpha - 2py \sin \alpha + 2p^2 - a^2 = 0$

The equation of a circle is $x^2 + y^2 = a^2$ and the equation of its chord is $x \cos \alpha + y \sin \alpha = p$. The equation of the circle for which this chord is a diameter is:

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