Recall the:
Distance Formula:
The distance between the points $(x_1,y_1)$ and $(x_2,y_2)$ is
$$
D=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}
$$
Example:
The radius of your circle is the distance between the points $(-1,4)$ and $(3,-2)$. Using the Distance Formula:
$$
D= \sqrt{ \bigl(3-(-1)\bigr)^2 + (-2-4)^2 }
= \sqrt{ 4^2 + (-6)^2 }
= \sqrt{ 16+36 }
= \sqrt{ 52 } .
$$
What is the equation of the circle?
It is important to realize the "equation of the circle" is: a point $(x,y)$ is on the circle if and only if the coordinates of the point $x$ and $y$ satisfy
the equation. So, how to get the equation? What is the relationship between the $x$ and $y$ coordinates of a point on the circle?
Well, let $(x,y)$ be a point on the circle. The big idea is:
$$
\text{The distance from the point }(x,y)\text{ to the center }(-1,4)\text{ is }\sqrt{52}.$$
So, using the distance formula (with $(x_2,y_2)=(x,y)$ and $(x_1,y_1)=(-1,4)$) , it follows that
$$
\sqrt{52}=\sqrt{\bigl(x-(-1)\bigr)^2 +(y-4)^2}.
$$
Or
$$
52=(x+1)^2 +(y-4)^2.
$$
The shortcut would be to just use the following formula (But it's important to realize why you'd use it and where it comes from):
Equation of a Circle
The equation of the circle with center located at $(a,b)$ and with radius $r$ is
$$
r^2=(x-a)^2 +(y-b)^2
$$
Note that the radius squared is on the left-hand side of the equation.
The following may help:
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.
![Circle.](https://i.stack.imgur.com/jml20.png)
Best Answer
You would rewrite this equation into the standard form for the equation of a circle. You would do this by completing the squares. For instance,
$x^2+y^2-4x+10y+26=0 \implies (x^2-4x+4)-4+(y^2+10y+25)-25+26=0 \implies (x-2)^2+(y+5)^2 = 3 \implies$
center of the circle is the point $(2,-5)$. Based on your insufficient information (no x-intercept is provided), there are infinitely many lines that pass through the center of the circle. We would have to know the x-intercept in order to solve this problem fully.