Circle Question

C is a circle with the equation x^2 + y^2 - 2x - 10y + 21.

The equation of this circle is (x-1)^2 + (y-5)^2 = 5

Given that the line joining P(3,6) and Q(q,4) is a diameter of C, show that q = -1.

So far, I've tried using Pythagoras Theorem to find q, since I know that the length of the diameter is √5.

Like so:

(3-q)^2 + (6-4)^2 = 2√5.

But this brings me to a quadratic, and two incorrect solutions.

What am I missing here?

Two things:

With the method you're using, your equation should be $(3-q)^2+(6-4)^2=(2\sqrt{5})^2$

However, since PQ is a diameter, P is a far to the right and up from the centre, as Q will be to the left and down. Similar triangles if you want to drawn in the detail. No quadratic required.

Thanks!

I've solved the question using a quadratic because that's the method I started doing it with.

Two more questions : Is q = -1 for one of the solutions of the quadratic enough to prove that q = -1?

Also, what is the other solution of the quadratic, 7?