Question

Find the equations to the circles which pass through the points $(5,7),(8,1)$, and $(1,3)$.

Answer 1

Three non-collinear points form a triangle, and a triangle will have only one circumcircle.
Let the center of the circle be $(h,k)$
Since the center is equidistant from all the points,
$(h-5{)}^2+(k-7{)}^2=(h-8{)}^2+(k-1{)}^2$ and
$(h-5{)}^2+(k-7{)}^2=(h-1{)}^2+(k-3{)}^2$
$\Rightarrow h^2-10h+25+k^2-14k+49=h^2-16h+64+k^2-2k+1$
i.e. $6h-12k+9=0$
i.e. $2h-4k+3=0..\left(1\right)$ and
$h^2-10h+25+k^2-14k+49=h^2-2h+1+k^2-6k+9$
i.e. $8h+8k-64=0$ or $h+k-8=0..\left(2\right)$
Multiplying equation $\left(2\right)$ by $4$ and adding that to equation $\left(1\right)$, we get
$6h-29=0$ or $h=\frac{29}{6}$
$\therefore k=8-\frac{29}{6}=\frac{19}{6}$
The radius thus becomes $\sqrt{{(8-\frac{29}{6})}^2+{(1-\frac{19}{6})}^2}=\sqrt{{\left(\frac{19}{6}\right)}^2+{\left(\frac{13}{6}\right)}^2}=\sqrt{\frac{361+169}{36}}=\sqrt{\frac{530}{36}}$
The equation of the circle becomes ${(x-\frac{29}{6})}^2+{(y-\frac{19}{6})}^2=\frac{265}{18}$