Finding an equation of circle which passes through three points How to find the equation of a circle which passes through these points $(5,10), (-5,0),(9,-6)$
using the formula 
$(x-q)^2 + (y-p)^2 = r^2$.
I know i need to use that formula but have no idea how to start, I have tried to start but don't think my answer is right.


I know i need to use that formula but have no idea how to start

\begin{equation*}
\left( x-q\right) ^{2}+\left( y-p\right) ^{2}=r^{2}\tag{0}
\end{equation*}
A possible very elementary way is to use this formula thrice, one for each point. Since the circle passes through the point $(5,10)$, it satisfies $(0)$, i.e.
$$\left( 5-q\right) ^{2}+\left( 10-p\right) ^{2}=r^{2}\tag{1}$$
Similarly for the second point $(-5,0)$:
$$\left( -5-q\right) ^{2}+\left( 0-p\right) ^{2}=r^{2},\tag{2}$$
and for $(9,-6)$:
$$\left( 9-q\right) ^{2}+\left( -6-p\right) ^{2}=r^{2}.\tag{3}$$
We thus have the following system of three simultaneous equations and in the three unknowns $p,q,r$:
$$\begin{cases}
\left( 5-q\right) ^{2}+\left( 10-p\right) ^{2}=r^{2} \\ 
\left( -5-q\right) ^{2}+p^{2}=r^{2} \\ 
\left( 9-q\right) ^{2}+\left( 6+p\right) ^{2}=r^{2}
\end{cases}\tag{4}
$$
To solve it, we can start by subtracting the second equation from the first  
$$\begin{cases}
\left( 5-q\right) ^{2}+\left( 10-p\right) ^{2}-\left( 5+q\right) ^{2}-p^{2}=0 \\ 
\left( 5+q\right) ^{2}+p^{2}=r^{2} \\ 
\left( 9-q\right) ^{2}+\left( 6+p\right) ^{2}=r^{2}
\end{cases}
$$
Expanding now the left hand side of the first equation we get a linear equation
$$\begin{cases}
100-20q-20p=0 \\ 
\left( 5+q\right) ^{2}+p^{2}=r^{2} \\ 
\left( 9-q\right) ^{2}+\left( 6+p\right) ^{2}=r^{2}
\end{cases}
$$
Solving the first equation for $q$ and substituting in the other equations, we get
$$\begin{cases}
q=5-p \\ 
\left( 10-p\right) ^{2}+p^{2}-\left( 4+p\right) ^{2}-\left( 6+p\right) ^{2}=0 \\ 
\left( 4+p\right) ^{2}+\left( 6+p\right) ^{2}=r^{2}
\end{cases}
$$
If we simplify the second equation, it becomes a linear equation in $p$ only
$$\begin{cases}
q=5-p \\ 
48-40p=0 \\ 
\left( 4+p\right) ^{2}+\left( 6+p\right) ^{2}=r^{2}
\end{cases}
$$
We have reduced our quadratic system $(4)$ to two linear equations plus the equation for $r^2$. From the second equation we find $p=6/5$, which we substitute in the first and in the third equations to find $q=19/5$ and $r^2=1972/25$, i.e
$$\begin{cases}
q=5-\frac{6}{5}=\frac{19}{5} \\ 
p=\frac{6}{5} \\ 
r^{2}=\left( 4+\frac{6}{5}\right) ^{2}+\left( 6+\frac{6}{5}\right) ^{2}=
\frac{1972}{25}.
\end{cases}\tag{5}
$$
So the equation of the circle is
\begin{equation*}
\left( x-\frac{19}{5}\right) ^{2}+\left( y-\frac{6}{5}\right) ^{2}=\frac{1972}{25}.
\end{equation*}
