Question

find the equation of the circle where (-9,4),(-2,5),(-8,-3),(-1,-2) are the vertices of an inscribed square.

Answer 1

Check the picture below, so, that'd be the square inscribed in the circle.

so... hmm the diagonals for the square are the diameter of the circle, and keep in mind that the radius of a circle is half the diameter, so let's find the diameter.

$\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -2}}\quad ,&{{ 5}})\quad % (c,d) &({{ -8}}\quad ,&{{ -3}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-8-(-2)]^2+[-3-5]^2} \\\\\\ d=\sqrt{(-8+2)^2+(-3-5)^2}\implies d=\sqrt{(-6)^2+(-8)^2} \\\\\\ d=\sqrt{36+64}\implies d=\sqrt{100}\implies d=10$

that means the radius r = 5.

now, what's the center?  well, the Midpoint of the diagonals, is really the center of the circle, let's check,

$\bf \textit{middle point of 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -2}}\quad ,&{{ 5}})\quad % (c,d) &({{ -8}}\quad ,&{{ -3}}) \end{array}\qquad \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right) \\\\\\ \left( \cfrac{-8-2}{2}~,~\cfrac{-3+5}{2} \right)\implies (-5~,~1)$

so, now we know the center coordinates and the radius, let's plug them in,

$\bf \textit{equation of a circle}\\\\ (x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2 \qquad \begin{array}{lllll} center\ (&{{ h}},&{{ k}})\qquad radius=&{{ r}}\\ &-5&1&5 \end{array} \\\\\\\ [x-(-5)]^2-[y-1]^2=5^2\implies (x+5)^2-(y-1)^2=25$