Question

Find the equation of the circle that has a diameter with endpoints located at (7, 3) and (7, –5).

Answer 1

So hmm check the picture below, that's about the circle and the endpoints, but notice, the endpoints make up a segment, namely the diameter of the circle, well.... let's see how long that is, because, the radius is half the diameter

$\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ 7}}\quad ,&{{ 3}})\quad % (c,d) &({{ 7}}\quad ,&{{ -5}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ d=\sqrt{(7-7)^2+(-5-3)^2}\implies d=\sqrt{0+(-8)^2}\implies d=8 \\\\\\ \textit{the radius is half that, so is }\boxed{r=4}$

now.. hmmm notice, the midpoint of the diameter, is the center of the circle, let's check that one out

$\bf \textit{middle point of 2 points }\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ 7}}\quad ,&{{ 3}})\quad % (c,d) &({{ 7}}\quad ,&{{ -5}}) \end{array}\qquad \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right) \\\\\\ \left(\cfrac{{{ 7+7}}}{2}\quad ,\quad \cfrac{{{ -5}} + {{ 3}}}{2} \right)\implies \left( \cfrac{14}{2}\ ,\ \cfrac{-2}{2} \right)\implies \boxed{(7,-1)}$

now.. that we know what the center is, and what the radius is, well

$\bf \textit{equation of a circle}\\\\ (x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2 \qquad \begin{array}{lllll} center\ (&{{ h}},&{{ k\quad }})\qquad radius=&{{ r}}\\ &7&-1&4 \end{array}$