<s>[INST] Find the equation of a circle given two points and a line that passes through its center Find the equation of the circle that passes through the points $(0,2)$ and $(6,6)$. Its center is on the line $x-y =1$.
 [/INST] First, find the center $(a, b)$.  It is equidistant from the two points, so
$$
\sqrt{a^2 + (b - 2)^2} = \sqrt{(a - 6)^2 + (b - 6)^2}
$$
or
$$
a^2 + (b - 2)^2 = (a - 6)^2 + (b - 6)^2.
$$
Now, since the center is on the line $x - y = 1$, we know that $a - b = 1$, or
$$
a = b + 1.
$$
Substitute this into the quadratic equation above and simplify:
$$
\begin{align}
(b + 1)^2 + (b - 2)^2 &= ((b + 1) - 6)^2 + (b - 6)^2 \\
(b^2 + 2b + 1) + (b^2 - 4b + 4) &= (b^2 - 10b + 25) + (b^2 - 12b + 36) \\
2b^2 - 2b + 5 &= 2b^2 - 22b + 61 \\
20b &= 56 \\
b &= \frac{14}{5}.
\end{align}
$$
Therefore,
$$
a = \frac{14}{5} + 1 = \frac{19}{5}.
$$
What is the radius $r$?  Well the square of the radius is
$$
\begin{align}
r^2 &= a^2 + (b - 2)^2 \\
&= \left( \frac{19}{5} \right)^2 + \left( \frac{4}{5} \right)^2 \\
&= \frac{377}{25}.
\end{align}
$$
Putting this together, we can write the equation of the circle as
$$
\left( x - \frac{19}{5} \right)^2 + \left( y - \frac{14}{5} \right)^2 = \frac{377}{25}
$$
or, by multiplying by $25$ to clear denominators,
$$
(5x - 19)^2 + (5y - 14)^2 = 377.
$$
Here's a picture.

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