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.

The green circle is the given circle defined by you. Let $K$ be the center of a circle through $A$ and tangent to the $y-\text{axis}$ at $J$. This is not necessarily your circle $C$. Let these two circles meet at $A$ and $L$, and let $M$ be the midpoint of common chord $AL$.

Now, suppose point $J$ moves freely along the $y-\text{axis}$. As $J$ travels, point $K$ is equidistant from point $A$ and the $y-\text{axis}$. The locus of $K$ is the parabola having focus $A$ and the $y-\text{axis}$ as directrix. Locus of $K$:
$y^2=-2a(x+\frac{a}{2})$
Again with point $J$ traveling, point $L$ traces circle $O$, while $A$ is stationary, and $M$ is the midpoint of $AL$. The locus of $M$ is the circle having diameter $AO$. Locus of $M$:
$x(x+a) + y^2 = 0$
The center of your circle $C$ must lie on both of these loci. Solve the system of equations. That gives you the coordinates of the circle center (two solutions). The rest should fall into place from there.
