The center of a circle is located at (−2, 7) . The radius of the circle is 2.

We have the form

( x - h)^2  + ( y - k)^2  = r^2      where  (h, k) is the center and r is the radius  ....so.....

(x + 2)^2  + (y - 7)^2   = 4           expand

x^2 + 4x + 4 + y^2 - 14y + 49  = 4       subtract 4 from both sides

x^2 + y^2 + 4x - 14y + 49 = 0

The center of a circle is located at (−2, 7) . The radius of the circle is 2.

What is the equation of the circle in general form?

A circle can be defined as the locus of all points that satisfy the equation
\((x-h)^2 + (y-k)^2 = r^2 \)  ( Standard Form )
where r is the radius of the circle,
and h,k are the coordinates of its center.

The general Form is:
\(x^2+y^2 +ax+by+c = 0\)

Standard Form to general Form:

\(\begin{array}{|rcll|} \hline (x-h)^2 + (y-k)^2 &=& r^2 \\ x^2-2xh+h^2+y^2-2yk+k^2 &=& r^2 \\ x^2+y^2+x\cdot\underbrace{(-2h)}_{=a}+y\cdot\underbrace{(-2k)}_{=b}+\underbrace{h^2+k^2-r^2}_{=c} &=& 0 \\ \hline \end{array} \)

a,b and c ?

\(\begin{array}{|rcll|} \hline x^2+y^2+x\cdot\underbrace{(-2h)}_{=a}+y\cdot\underbrace{(-2k)}_{=b}+\underbrace{h^2+k^2-r^2}_{=c} &=& 0 \\\\ \color{red}a &\color{red}=& \color{red}-2h \\\\ \color{red}b &\color{red}=& \color{red}-2k \\\\ \color{red}c &\color{red}=&\color{red}h^2+k^2-r^2\\ \hline \end{array} \)

If we have h,k and r, we can calculate a,b and c:

\(\begin{array}{|lcll|} \hline \mathbf{x^2+y^2 +ax+by+c = 0} \\ a = -2h \\ b = -2k \\ c =h^2+k^2-r^2 \\ \hline \end{array}\)

\(h=-2\\ k=7\\ r=2\)

\(\begin{array}{|lcll|} \hline a = -2h \\ a = -2\cdot(-2)\\ \mathbf{a = 4} \\\\ b = -2k \\ b = -2(7) \\ \mathbf{a = -14} \\\\ c =h^2+k^2-r^2 \\ c =(-2)^2+7^2-2^2 \\ c =4+49-4 \\ \mathbf{c =49} \\\\ x^2+y^2 +ax+by+c = 0 \\ \mathbf{x^2+y^2 +4x-14y+49 =0} \\ \hline \end{array} \)

The equation of the circle in general form is: \(x^2+y^2 +4x-14y+49 =0\)