If the straight line x cos alpha + y sin alph | vedclass

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(A) The family of circles passing through the intersection of the circle $S: x^2 + y^2 - a^2 = 0$ and the line $L: x \cos \alpha + y \sin \alpha - P =

Vedclass · Accepted Answer

$x^2 + y^2 - 2Px \cos \alpha - 2Py \sin \alpha + 2P^2 - a^2 = 0$

Vedclass · Answer

(A) The family of circles passing through the intersection of the circle $S: x^2 + y^2 - a^2 = 0$ and the line $L: x \cos \alpha + y \sin \alpha - P = 0$ is given by $S + \lambda L = 0$.$x^2 + y^2 - a^2 + \lambda(x \cos \alpha + y \sin \alpha - P) = 0$$x^2 + y^2 + \lambda x \cos \alpha + \lambda y \sin \alpha - a^2 - \lambda P = 0$ $(i)$The centre of this circle is $\left(-\frac{\lambda \cos \alpha}{2}, -\frac{\lambda \sin \alpha}{2}\right)$.Since the line $x \cos \alpha + y \sin \alpha = P$ is the diameter $\overline{AB}$,the centre of the circle must lie on this line.Substituting the centre into the line equation:$\left(-\frac{\lambda \cos \alpha}{2}\right) \cos \alpha + \left(-\frac{\lambda \sin \alpha}{2}\right) \sin \alpha = P$$-\frac{\lambda}{2} (\cos^2 \alpha + \sin^2 \alpha) = P$$-\frac{\lambda}{2} = P \Rightarrow \lambda = -2P$Substituting $\lambda = -2P$ into equation $(i)$:$x^2 + y^2 - 2Px \cos \alpha - 2Py \sin \alpha - a^2 - (-2P)P = 0$$x^2 + y^2 - 2Px \cos \alpha - 2Py \sin \alpha + 2P^2 - a^2 = 0$

If the straight line $x \cos \alpha + y \sin \alpha = P$ intersects the circle $x^2 + y^2 = a^2$ at $A$ and $B$,then the equation of the circle with diameter $\overline{AB}$ is

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