Finding an equation of circle which passes through three points

Question

How to find the equation of a circle which passes through these points $(5,10), (-5,0),(9,-6)$ using the formula $(x-q)^2 + (y-p)^2 = r^2$.

I know i need to use that formula but have no idea how to start, I have tried to start but don't think my answer is right.

Answer 1

I know i need to use that formula but have no idea how to start

\begin{equation*} \left( x-q\right) ^{2}+\left( y-p\right) ^{2}=r^{2}\tag{0} \end{equation*}

A possible very elementary way is to use this formula thrice, one for each point. Since the circle passes through the point $(5,10)$, it satisfies $(0)$, i.e.

$$\left( 5-q\right) ^{2}+\left( 10-p\right) ^{2}=r^{2}\tag{1}$$

Similarly for the second point $(-5,0)$:

$$\left( -5-q\right) ^{2}+\left( 0-p\right) ^{2}=r^{2},\tag{2}$$

and for $(9,-6)$:

$$\left( 9-q\right) ^{2}+\left( -6-p\right) ^{2}=r^{2}.\tag{3}$$

We thus have the following system of three simultaneous equations and in the three unknowns $p,q,r$:

$$\begin{cases} \left( 5-q\right) ^{2}+\left( 10-p\right) ^{2}=r^{2} \\ \left( -5-q\right) ^{2}+p^{2}=r^{2} \\ \left( 9-q\right) ^{2}+\left( 6+p\right) ^{2}=r^{2} \end{cases}\tag{4} $$

To solve it, we can start by subtracting the second equation from the first

$$\begin{cases} \left( 5-q\right) ^{2}+\left( 10-p\right) ^{2}-\left( 5+q\right) ^{2}-p^{2}=0 \\ \left( 5+q\right) ^{2}+p^{2}=r^{2} \\ \left( 9-q\right) ^{2}+\left( 6+p\right) ^{2}=r^{2} \end{cases} $$

Expanding now the left hand side of the first equation we get a linear equation

$$\begin{cases} 100-20q-20p=0 \\ \left( 5+q\right) ^{2}+p^{2}=r^{2} \\ \left( 9-q\right) ^{2}+\left( 6+p\right) ^{2}=r^{2} \end{cases} $$

Solving the first equation for $q$ and substituting in the other equations, we get

$$\begin{cases} q=5-p \\ \left( 10-p\right) ^{2}+p^{2}-\left( 4+p\right) ^{2}-\left( 6+p\right) ^{2}=0 \\ \left( 4+p\right) ^{2}+\left( 6+p\right) ^{2}=r^{2} \end{cases} $$

If we simplify the second equation, it becomes a linear equation in $p$ only

$$\begin{cases} q=5-p \\ 48-40p=0 \\ \left( 4+p\right) ^{2}+\left( 6+p\right) ^{2}=r^{2} \end{cases} $$

We have reduced our quadratic system $(4)$ to two linear equations plus the equation for $r^2$. From the second equation we find $p=6/5$, which we substitute in the first and in the third equations to find $q=19/5$ and $r^2=1972/25$, i.e

$$\begin{cases} q=5-\frac{6}{5}=\frac{19}{5} \\ p=\frac{6}{5} \\ r^{2}=\left( 4+\frac{6}{5}\right) ^{2}+\left( 6+\frac{6}{5}\right) ^{2}= \frac{1972}{25}. \end{cases}\tag{5} $$

So the equation of the circle is

\begin{equation*} \left( x-\frac{19}{5}\right) ^{2}+\left( y-\frac{6}{5}\right) ^{2}=\frac{1972}{25}. \end{equation*}

Answer 2

if the problem has a solution (the three points are on a circle) then you only need to calculate the equations of the two mediators and the intersection should be the center and the distance between one of the points and this center gives you r.

Answer 3

Suppose the points are $A,B,C$. Then intersect the equations of perpendicular bisectors of $AB$ and $BC$. This is the center of the desired circle. (with your notation $(p,q)$)

Now calculate the distance between $(p,q)$ and $A$. Now $r$ is also found.

Answer 4

$\begin{vmatrix} x^2+y^2&x&y&1\\ 5^2+10^2&5&10&1\\ (-5)^2+0^2&-5&0&1\\ 9^2+(-6)^2&9&-6&1\\ \end{vmatrix}= \begin{vmatrix} x^2+y^2&x&y&1\\ 125&5&10&1\\ 25&-5&0&1\\ 117&9&-6&1\\ \end{vmatrix} = 0$

Answer 5

This method that I wrote creates a circle object which has a radius and a centre point. (written in Swift but can be port to many other languages.)

  func circle(p1: (x: Float,y: Float), p2: (x: Float,y: Float), p3: (x: Float,y: Float)) -> (r: Float, c: (x: Float,y: Float)) {
    let x1 = p1.x
    let x2 = p2.x
    let x3 = p3.x
    let y1 = p1.y
    let y2 = p2.y
    let y3 = p3.y

    let mr = (y2-y1) / (x2-x1);
    let mt = (y3-y2) / (x3-x2);

    if mr == mt { return nil }

    let x12 = x1 - x2;
    let x13 = x1 - x3;

    let y12 = y1 - y2;
    let y13 = y1 - y3;

    let y31 = y3 - y1;
    let y21 = y2 - y1;

    let x31 = x3 - x1;
    let x21 = x2 - x1;

    // x1^2 - x3^2
    let sx13 = pow(x1, 2) -
      pow(x3, 2);

    // y1^2 - y3^2
    let sy13 = pow(y1, 2) -
      pow(y3, 2)

    let sx21 = pow(x2, 2) -
      pow(x1, 2)

    let sy21 = pow(y2, 2) -
      pow(y1, 2)

    let f = ((sx13) * (x12)
      + (sy13) * (x12)
      + (sx21) * (x13)
      + (sy21) * (x13))
      / (2 * ((y31) * (x12) - (y21) * (x13)))
    let g = ((sx13) * (y12)
      + (sy13) * (y12)
      + (sx21) * (y13)
      + (sy21) * (y13))
      / (2 * ((x31) * (y12) - (x21) * (y13)))

    let c = -pow(x1, 2) - pow(y1, 2) -
      2 * g * x1 - 2 * f * y1;

    let h = -g;
    let k = -f;
    let sqr_of_r = h * h + k * k - c;

    let r = sqrt(sqr_of_r)

    return (r: r, center: (x: h, y: k))
  }