7
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ Can you share what you've tried, and explain what you're having trouble with? You have three data points, and three variables to solve for. $\endgroup$
    – user61527
    Jun 8, 2014 at 17:48
  • 1
    $\begingroup$ So far i have the following but it just doesn't seem to sit right with me as being correct Sqareroot[(9-5-5)^2 + (-6-0-10)^2] and that would be the radius $\endgroup$
    – help
    Jun 8, 2014 at 18:14

5 Answers 5

17
$\begingroup$

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.

circle

$\endgroup$
12
$\begingroup$

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

$\endgroup$
4
  • 2
    $\begingroup$ The simplest answer. $\endgroup$
    – user5402
    Jun 8, 2014 at 20:12
  • 7
    $\begingroup$ what does it say ? $\endgroup$ Jun 8, 2014 at 21:00
  • 1
    $\begingroup$ I don't get it? $\endgroup$
    – Vrisk
    Oct 26, 2017 at 13:59
  • $\begingroup$ How to get the matrix form? $\endgroup$ Feb 3, 2021 at 9:49
6
$\begingroup$

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*}

$\endgroup$
4
$\begingroup$

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))
  }

$\endgroup$
5
  • $\begingroup$ Are you sure that your solution covers all degenerate cases ? $\endgroup$
    – ogerard
    May 5, 2015 at 14:54
  • $\begingroup$ Yes I am sure. If you encounter any, please let me know. $\endgroup$
    – Erk Ekin
    Feb 9, 2017 at 8:16
  • $\begingroup$ If y2 = y1, then mr will be 0, and y will fail with division by 0. You need more checks to see if the points sit on one line, and to avoid false positives. The solution (if there is in the case of 3 given points) is to rotate them. $\endgroup$
    – Andris
    Apr 14, 2018 at 22:49
  • $\begingroup$ Nice example! Works just fine. Aditional test are easy to include. In Java, you can use vecmath lib (javax.vecmath). $\endgroup$
    – Alex Byrth
    May 2, 2018 at 18:41
  • $\begingroup$ I noticed @Andris' point and fixed it. I didn't rotate but applied matrix solution instead. $\endgroup$
    – Erk Ekin
    Jan 11, 2020 at 17:29
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Perhaps you could show how? At least show some guidelines. $\endgroup$
    – user88595
    Jun 8, 2014 at 18:19
  • $\begingroup$ This describes the Erk Ekin above implementation. $\endgroup$
    – Alex Byrth
    May 2, 2018 at 18:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.