Question

Write the general equation for the circle that passes through the points:

Answer 1

Check the picture below.

since the third point there, is more or less between the other two, chances are is a radius distance from the center, namely, other two points you see in the picture, are pretty much the diameter of the circle, and half that distance is the radius, and their midpoint is the center of the circle.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{1}~,~\stackrel{y_1}{7})\qquad (\stackrel{x_2}{7}~,~\stackrel{y_2}{-1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{(7-1)^2+(-1-7)^2}\implies d=\sqrt{6^2+(-8)^2} \\\\\\ d=\sqrt{36+64}\implies d=\sqrt{100}\implies d=10~\hfill \stackrel{\textit{half of that or radius}}{r = 5} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{1}~,~\stackrel{y_1}{7})\qquad (\stackrel{x_2}{7}~,~\stackrel{y_2}{-1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{7+1}{2}~~,~~\cfrac{-1+7}{2} \right)\implies \left( \cfrac{8}{2}~,~\cfrac{6}{2} \right)\qquad \implies \qquad \stackrel{center}{(4,3)} \\\\[-0.35em] ~\dotfill$

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{4}{ h},\stackrel{3}{ k})\qquad \qquad radius=\stackrel{5}{ r} \\\\\\ (x-4)^2+(y-3)^2=5^2\implies (x-4)^2+(y-3)^2=25 \\\\\\ \stackrel{\mathbb{F~O~I~L}}{(x^2-8x+16)}+\stackrel{\mathbb{F~O~I~L}}{(y^2-6y+9)}=25\implies x^2-8x+y^2-6y+25=25 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill x^2+y^2-8x-6y=0~\hfill$