Find the equation of a circle passing through 3 points

Question

Find the equation of a circle which passes through these points: (2,-1) , (-2,3) , (1,5)

Answer 1

Circle-equation is $(x-p)^2+(y-q)^2=r^2$, where $(p,q)$ is center and $r$ radius. We have to solve system $$(2-p)^2+(1+q)^2=r^2,$$ $$(2+p)^2+(3-q)^2=r^2,$$ $$(1-p)^2+(5-q)^2=r^2,$$ which is equivalent with (system (1),(2)-(1),(3)-(1))$$(2-p)^2+(1+q)^2=r^2,$$ $$8p+8-8q=0,$$ $$21-6p-12q=0.$$ From two last equations we get $p=\frac{1}{2}$, $q=\frac{3}{2}$ and now substitute into first equation to get $r=\frac{\sqrt{34}}{2}.$

Answer 2

Hint: A nice way to find the center of the circle is to find the intersection of two perpendicular bisectors.

Answer 3

Let $O(a,b)$ be the centre of the circle with radius $r$ . Let $A(2,-1)$ , $B(-2,3)$ and $C(1,5)$ .

Then You need to solve following system of equations :

$$\begin{cases} (x_A-a)^2+(y_A-b)^2=r^2 \\ (x_B-a)^2+(y_B-b)^2=r^2\\ (x_C-a)^2+(y_C-b)^2=r^2 \end{cases}$$