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Get the equation of a circle through the points $(1,1), (2,4), (5,3) $.

I can solve this by simply drawing it, but is there a way of solving it (easily) without having to draw?

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Hint: Find the center of the circle. –  ᴊ ᴀ s ᴏ ɴ Oct 14 '12 at 15:09
How would one do that? –  JohnPhteven Oct 14 '12 at 15:10
@ZafarS: the center lies on the perpendicular bisector of every line segment between two points on the circle. –  Henning Makholm Oct 14 '12 at 15:13
So I must basically find for examply the bisector of AC and BC and where the line crosses is the M? –  JohnPhteven Oct 14 '12 at 15:17
@ Henning Makholm: Your method is too complicated, we can simply set a coordinate for a point that has equal distance from each point. –  ᴊ ᴀ s ᴏ ɴ Oct 15 '12 at 8:20

5 Answers 5

up vote 3 down vote accepted

Big hint:

Let $A\equiv (1,1)$,$B\equiv (2,4)$ and $C\equiv (5,3)$.

We know that the perpendicular bisectors of the three sides of a triangle are concurrent.Join $A$ and $B$ and also $B$ and $C$.

The perpendicular bisector of $AB$ must pass through the point $(\frac{1+2}{2},\frac{1+4}{2})$

Now find the equations of the straight lines AB and BC and after that the equation of the perpendicular bisectors of $AB$ and $BC$.Solve for the equations of the perpendicular bisectors of $AB$ and $BC$ to get the centre of your circle.

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Ok, this is my answer: Line segment AB has the formula $AB = 0,5x+0,5$ , so its bisector has $y=-2x+8$ and Line segment AC has formula $AC=3x-2$, so its bisector has $y=-\dfrac{1}{3}x+3$. Look for the intersection between the lines: $(3,2)$. From there it's easy to find that $r=\sqrt{5}$, so $(x-3)^2+(y-2)^2 = 5$ –  JohnPhteven Oct 14 '12 at 15:26
You seem to be on the right track.Write up your answer and post it as an answer to get feedback! –  Richard Nash Oct 14 '12 at 15:28

Follow these steps:

  1. Consider the general equation for a circle as $(x-x_c)^2+(y-y_c)^2 - r^2 = 0$
  2. Plug in the three points to create three quadratic equations $$ (1-x_c)^2+(1-y_c)^2 - r^2 = 0 $$ $$ (2-x_c)^2+(4-y_c)^2 - r^2 = 0 $$ $$ (5-x_c)^2+(3-y_c)^2 - r^2 = 0 $$
  3. Subtract the first from the second, and the first from the third to create two linear equations $$ -2 x_c -6 (y_c-3)=0 $$ $$ (y_c+7)-6 x_c = 0 $$
  4. Solve for the center as $$ (x_c,y_c) = (3,2) $$
  5. Plug the values for the center in any of the three quadratic equations above (I choose the first) and solve for $r$ $$ (1-3)^2+(1-2)^2-r^2 = 0 $$ $$ 5-r^2 = 0 $$ $$ r = \sqrt 5 $$
  6. Verify result with GeoGebra (optional)


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Anytime you turn a non-linear problem into a linear problem it is a win. –  ja72 Apr 28 '14 at 20:54

Consider the general (implicit) equation that defines a circle, with parameters $\alpha$, $\beta$, $\gamma$. Substitute the coordinate of the given points and get three linear equations in the three variables $\alpha$, $\beta$, $\gamma$. Solve the system.

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You can also find first $R$ from the sin Law:

$$R= \frac{BC}{2 \sin (A)}= \frac{BC \cdot AB \cdot AC }{2 \| AB \times AC \|} \tag{$*$}$$

Next, write the equations of circles of radius $R$ with centre $A$ and $B$ and solve.

Note The formula $(*)$ is the well known geometric formula for the area of a triangle:

$$\mbox{Area}= \frac{abc}{4R} \,.$$

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This was just what I was looking for. This kind of formula was quick to implement. stackoverflow.com/a/26903599/999943 –  phyatt Nov 13 '14 at 7:51

I'm surprised this one hasn't been mentioned; you can find the equation by using the determinant of a matrix:

$$\left|\begin{array}{cccc} x^2+y^2&x&y&1\\ 1^2+1^2&1&1&1\\ 2^2+4^2&2&4&1\\ 5^2+3^2&5&3&1\\ \end{array}\right|=0$$ This gives the equation of the circle through those three points. This sort of thing can be used in a lot of situations: matrix-determinant solutions are available for any shape I can think of where you're given points that land on the shape.

Implementing this on a computer involves having a thing that calculates determinants; to do it numerically you'll need to apply the cofactors method to avoid multiplying the variables in.

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One of my favorite tricks - extends to more general conics as well! –  Mark McClure Apr 7 at 22:03

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