how do you find the equation of a circle if the ends of the diameter are (18,-13) and (4,-3)​

The equation of a circle if the ends of the diameter are (18,-13) and (4,-3)​ is [tex](x-11)^{2}+(y+8)^{2}=74[/tex]Solution:Given that ends of diameter are (18, -13) and (4, -3)To find the equation of circle, let us first find the center (h, k) of the circleWe know that the center of the circle lies in the center of the diameter also. In order to find the center, get the average of both x and y[tex]\mathrm{c}=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)[/tex][tex]\begin{array}{l}{c=\frac{18+4}{2}, \frac{-13-3}{2}} \\\\ {c=\frac{22}{2}, \frac{-16}{2}} \\\\ {\mathrm{c}=(11,-8)}\end{array}[/tex]These are the coordinates of the center of the circle Now we need to find the radius, we need the center (h, k) and one of the given points (4, -3): The equation of circle is given as:[tex](x-h)^{2}+(y-k)^{2}=r^{2}[/tex][tex]\begin{array}{l}{(4-11)^{2}+(-3-(-8))^{2}=r^{2}} \\\\ {(-7)^{2}+(-3+8)^{2}=r^{2}} \\\\ {(-7)^{2}+(5)^{2}=r^{2}} \\\\ {49+25=r^{2}} \\\\ {r=\sqrt{74}}\end{array}[/tex]So the equation of the circle is:[tex]\begin{array}{l}{(x-11)^{2}+(y-(-8))^{2}=(\sqrt{74})^{2}} \\\\ {(x-11)^{2}+(y+8)^{2}=74}\end{array}[/tex]