find the equation of a circle from 3 points on circumference The following question is from higher maths 2014 Scotland

(a) Find P and Q, the points of intersection of the line ${y = 3x -
 5}$ and the circle ${C_1}$ with the equation ${x^2 + y^2 + 2x - 4y -15
 =0}$.
(b) T is at the centre ${C_1}$.  Show PT and QT are perpendicular.
(c) A second circle  ${C_2}$ passes through P, Q and T.  Find the
  equation of  ${C_2}$.

I have solved a and b but I am stuck on part (c).
After parts a and b, I have 3 points on the circumference P (1, -2), Q (3, 4) and T (-1, 2).
I know that a perpendicular bisector to a chord will always pass through the centre.
So am I right in saying that I should find the perpendicular bisectors of PT and QT (I have the gradients from question b) and then solve the 2 equations simultaneously.
The perpendicular bisector of PT is
Midpoint PT = ${({1 + 1 \over 2}, {-2 + 2 \over 2})}$ = (0, 0)
gradient of PT = ${2 - -2 \over -1 - 1}$ = -2, perpendicular gradient = ${1 \over 2}$
equation of PT = ${y = 1\over2 x}$
=> ${2y -x = 0}$
Midpoint of QT = ${({3 + -1\over 2}, {2 + 4 \over 2})}$ = (1, 3)
gradient of QT = ${-2}$
equation of QT = ${y - 2 = -2(x + 1)}$
=> ${y - 2 = -2x -2}$
=> ${y + 2x = 0}$
I would then solve them simultaneously to find the point (0, 0) which seems wrong.
I take it I have taken a wrong term somewhere?
 A: You have a mistake in finding the perpendicular bisector of $QT$.
The equation of the line, whose slope is $-2$, passing through $(1,3)$ is
$$y-3=-2(x-1),$$
not
$$y-2=-2(x+1).$$
A: mathlove has pointed to your error, but here is a simpler way to proceed:
You proved in part (b) that $PT$ and $QT$ are perpendicular, so $PTQ$ is a right triangle, and the center of its circumcircle is the midpoint of the hypotenuse $PQ$.
(It would also be simple to see this by plotting the points on graph paper!)
A: An easy solution is achieved by translating the point $P$ to the origin (and $Q,T$ accordingly).
Then the equation of a circle by the origin is
$$ax+by=x^2+y^2.$$
Plugging the coordinates of $Q-P$ and $T-P$, you easily solve
$$ax_q+by_q=x_q^2+y_q^2\\
ax_t+by_t=x_t^2+y_t^2,$$
and the coordinates of the center are $\frac12(a,b)+P$, and the radius $\frac12\sqrt{a^2+b^2}$.
A: First, put $C_1$ in standard form:
$$(x+1)^2+(y-2)^2=20$$
a) is algebra, easy to verify that $(3,4)$ and $(1, -2)$ are the two solutions.
b) is easiest, I think, by showing that the triangle $\Delta TPQ$ is $45-45-90$, which requires showing that $\overline{PQ}$ is $\sqrt{2}$ as long as the radius of $C_1$, i.e. $\sqrt{40}$, which it is (easy application of distance formula).
I think c) is easy because we know the circumcircle of a right triangle has its center on the midpoint of the hypotenuse (easy to see that's $(2, 1)$), and thus radius half the hypotenuse ($\frac{\sqrt{40}}2$). Which, in standard form, is:
$$(x-2)^2+(y-1)^2=10$$
(and blessedly these are all integers, so it's easy to verify that $T$, $P$, and $Q$ all satisfy this)
A: In general, the equation of the circle through the three points $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is 
$$
\left| \;
\begin{matrix}
x^2 + y^2     &  x   & y   & 1 \\
x_1^2 + y_1^2 &  x_1 & y_1 & 1 \\ 
x_2^2 + y_2^2 &  x_2 & y_2 & 1 \\
x_3^2 + y_3^2 &  x_3 & y_3 & 1 \\  
\end{matrix}
\;\right| = 0
$$
