What is the equation of the circle I've been experiencing a difficulties in answering this. I hope someone will help me in solving this
Find the equation of the circle through the points $(2,8),(7,3)$ and $(-2,0)$.
 A: Let $A(2,8),B(7,3)$ and $C(-2,0)$
Using, "$\perp$ line from centre bisects chord"
Midpoint of AC is $M_{AC}=(0,4)$
Midpoint of BC is $M_{BC}=\left(\dfrac{5}{2},\dfrac{3}{2}\right)$
Slope of line the lines AC and BC are $m_{AC}=2$, $m_{BC}=\frac{1}{3}$
and slope of perpendicular bisectors of AC and BC $m_1=-\frac{1}{2}$,$m_2=-3$,
and equation of perpendicular bisectors are $y=-\frac{1}{2}x+4$ and  $y=-3x+9$, respectively
Coordinates of intersection of perpendicular bisectors are $O(2;3)$ that is centre of the circle.
Radius is $r=OC=5$ or $r=OA=OB=OC$
$(x-2)^2+(y-3)^2=25$
A: HINT
Let start from the general equation
$$(x-x_C)^2+(y-y_C)^2=R^2$$
and plug in the values $(x,y)$ for the given points to obtain three equations in the three uknowns $x_C$, $y_C$ and $R$.
We obtain the system

*

*$2x_1x_C+2y_1y_C-x_1^2-y_1^2=x_C^2+y_C^2-R^2$

*$2x_2x_C+2y_2y_C-x_2^2-y_2^2=x_C^2+y_C^2-R^2$

*$2x_3x_C+2y_3y_C-x_3^2-y_3^2=x_C^2+y_C^2-R^2$
that is

*

*$2x_1x_C+2y_1y_C-x_1^2-y_1^2=2x_2x_C+2y_2y_C-x_2^2-y_2^2$

*$2x_2x_C+2y_2y_C-x_2^2-y_2^2=2x_3x_C+2y_3y_C-x_3^2-y_3^2$
from which we can find $x_C$ and $y_C$ and then $R$.

We can check directly the solution $(x-2)^2+(y-3)^2=25$ you have obtaind as follows

*

*$(2,8) \to (2-2)^2+(8-3)^2=25$

*$(7,3) \to (2-7)^2+(3-3)^2=25$

*$(-2,0) \to (-2-2)^2+(0-3)^2=25$
A: 
The question can be restated as

Given triangle $ABC$ with coordinates of vertices $A=(-2,0)$,
$B=(7,3)$, $C=(2,8)$, find the equation of its circumscribed circle.

First, find the squares of the side lengths of $\triangle ABC$:
\begin{align}
a^2&=50
,\quad
b^2= 80
,\quad
c^2=90
\tag{1}\label{1}
.
\end{align}
Second, the coordinates of the circumcenter
is known to be found as
\begin{align}
O&=
\frac{a^2(b^2+c^2-a^2)\cdot A+b^2(a^2+c^2-b^2)\cdot B+c^2(b^2+a^2-c^2)\cdot C}
{a^2(b^2+c^2-a^2)+b^2(a^2+c^2-b^2)+c^2(b^2+a^2-c^2)}
\\
&=\tfrac1{12}\cdot(5\,A+4\,B+3\,C)
=\tfrac1{12}\cdot(5\,(-2,0)+4\,(7,3)+3\,(2,8))
=(2,3)
\tag{2}\label{2}
.
\end{align}
Third, find the radius:
\begin{align}
R&=|O-A|=|O-B|=|O-C|=5
\tag{3}\label{3}
.
\end{align}
Hence, the equation of the circle is
\begin{align}
(x-2)^2+(y-3)^2&=25
\tag{4}\label{4}
.
\end{align}
A: Satisfy the general equation of circle
$$x^2+y^2+2gx+2fy+c=0$$
with these  three points get linear equations in $f,g,c$ and solve them.
