Find the equation of a circle whose diameter is the chord of another circle I am studying maths as a hobby and have just come to the last end-of-chapter question on parabolas and circles.
The straight line through the point $A(-a,0)$ at an angle $\theta$ to the positive direction of the x-axis meets the circle $x^2+y^2=a^2$ at P, distinct from A. The circle on AP as diameter is denoted by C.
i) By finding the equation of C, or otherwise, show that if C touches the y-axis, then $\cos2\theta=2-\sqrt 5$
ii) C meets the x-axis at M, distinct from A, and the tangents to C at A and M meet at Q. Find the coordinates of Q in terms of $\theta$ and show that as $\theta$ varies, Q always lies on the curve $y^2x +(x+a)^3=0$.
I really don't know how to begin here. To find the equation of C I am thinking I need two points so I can use the equation: $$x^2+y^2+2gx+2fy+c=0$$
But I only have one definite point, at A. The value of y at B is unknown. I could call P (x,y) but this is too vague.
This is my visualisation of the situation for the first part:

I can see that $\angle ABP$ is a right angle.

A simpler observation is to note that the coordinate of $P$ is $(a \cos 2\theta, a \sin 2\theta)$, since the central angle that $OP$ makes with the positive $x$-axis is twice the corresponding inscribed angle $\angle PAO$.
Moreover, because $\triangle AOP$ is isosceles, the midpoint $D = \frac{a}{2}(-1 + \cos 2\theta, \sin 2\theta)$ of $AP$ is the foot of the altitude $DO$, hence $\triangle AOD$ implies $AD = a \cos \theta$.
But since $BD = AD$, it follows that $$\frac{a}{2} (1 - \cos 2\theta) = a \cos \theta = a \sqrt{\frac{1 + \cos 2\theta}{2}}.$$  Letting $z = \cos 2\theta$, we obtain a quadratic in $z$ after cancelling $a$:
$$(1-z)^2 = 2 (1 + z),$$ or $$0 = z^2 - 4z - 1,$$ hence $\cos 2\theta \in \left\{ 2 \pm \sqrt{5} \right\}$.  Since only one of these roots has magnitude not exceeding $1$, we obtain $\cos 2\theta = 2 - \sqrt{5}$ as claimed.
