Find the equation of a circle containing three given points

Question

Problem:

Determine the equation of the circle that passes through three points, $J(-3, 2)$, $K(4, 1)$, and $L(6, 5)$.

I thought of using systems like so: $$\left\{ \begin{array}{rcl} (x+3)^2 + (y-2)^2 = r^2 \\ (x-4)^2 + (y-1)^2 = r^2 \\ (x-6)^2 + (y-5)^2 = r^2 \\ \end{array} \right.$$ After equating the expressions on the left hand side of each equation, expanding and simplifying, I found out that $y = 7x -2$. I decided to substitute the $y$ into two of the expressions to solve for $x$. But it only gives me $50x^2 - 50x + 25 = 50x^2 - 50x + 25$, which does not help. Can someone please help me out? Thanks!

Answer 1

Your procedure (modified a little) will work. From the first two equations, you got a linear equation in $x$ and $y$. In the same way, from the last two equations, you can get a linear equation in $x$ and $y$. Solve. Now you have the coordinates of the centre, and the rest is easy.

Remark: The procedure will be clearer if you start by saying let $(a,b)$ be the centre, and $r$ the radius. Then write down your equations, but with $a,b$ instead of $x,y$. These equations say that the three given points are all at distance $r$ from the centre.

Answer 2

Use the equation $x^2+y^2+2gx+2fy+c=0$. Substitute those values, you will get three equations involving g,f,c. Substituting the values back to the equation will be your equation. The answer will be $x^2+y^2-2x-10y+1=0$ or $(x-1)^2+(y-5)^2=25$

Answer 3

Hints following Gerry's comment (=huge hint), a much easier and elementary (in my opinion, of course) method:

$$\begin{align*}\text{Middle point of segment}\;\;JK:&\;\;\left(\frac12\,,\,\frac32\right)\\{}\\ \text{Middle point of segment}\;\;JL:&\;\;\left(\frac32\,,\,\frac72\right)\\{}\\ \text{Middle point of segment}\;\;KL:&\;\;\;\;\left(5\,,\,3\right)\end{align*}$$

Now calculate two of the bisectrices of the above segments, for example the bisectrix of segment $\;JK\;$:

$$\text{Slope of JK}:\;\;m_{JK}=-\frac17\implies\;\text{slope of bisectrix}\;\;m=7\implies$$

$$\text{equation of bisectrix}:\;\;y-\frac32=7\left(x-\frac12\right)\implies y=7x-2$$

After that, find intersection of these two bisectrices: this point is the the circle's center (why?), and then just find the distance from any of the three given points to the center to know what's the circle's center and you're done.

Answer 4

$\begin{vmatrix} x^2+y^2&x&y&1\\ (-3)^2+2^2&-3&2&1\\ 4^2+1^2&4&1&1\\ 6^2+5^2&6&5&1\\ \end{vmatrix}=0$

$30(x^2+y^2)-60x-300y+30=0$

$x^2+y^2-2x-10y+1=0$