Answer
Center of circle is $\left( -\frac{7}{2},\frac{3}{2} \right)$ and radius is $r=\frac{\sqrt{98}}{2}$
Work Step by Step
Standard equation of the circle is:
${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ (equation - 1)
And equation of circle is ${{x}^{2}}+{{y}^{2}}+7x-3y=10$ (equation - 2)
Now add $\frac{49}{4}$ and $\frac{9}{4}$ on both the sides of (equation - 2) to complete the square twice.
$\begin{align}
& {{x}^{2}}+{{y}^{2}}+7x-3y+\frac{49}{4}+\frac{9}{4}=10+\frac{49}{4}+\frac{9}{4} \\
& \left( {{x}^{2}}+7x+\frac{49}{4} \right)+\left( {{y}^{2}}-3y+\frac{9}{4} \right)=\frac{40+49+9}{4} \\
& {{\left( x+\frac{7}{2} \right)}^{2}}+{{\left( y-\frac{3}{2} \right)}^{2}}=\frac{98}{4} \\
& {{\left( x+\frac{7}{2} \right)}^{2}}+{{\left( y-\frac{3}{2} \right)}^{2}}={{\left( \frac{\sqrt{98}}{2} \right)}^{2}}
\end{align}$
Now compare the standard equation with the equation${{\left( x+\frac{7}{2} \right)}^{2}}+{{\left( y-\frac{3}{2} \right)}^{2}}={{\left( 7 \right)}^{2}}$.
Center coordinate of circle is $\left( h=-\frac{7}{2},k=\frac{3}{2} \right)$.
And radius of circle is $r=\frac{\sqrt{98}}{2}$.
To graph, we plot the points $\left( -3.5,6.419 \right)$, $\left( -3.5,-3.419 \right)$, $\left( -8.42,1.5 \right)$, and $\left( 1.42,1.5 \right)$ which are, respectively, $\frac{\sqrt{98}}{2}$ units above, below, left and right of $\left( -\frac{7}{2},\frac{3}{2} \right)$.
