Finding the equation of a circle given three points on the circle.

Question

Problem:

Find an equation for the circle though the points $(2,3)$, $(3,2)$ and $(-4,3)$ Answer:

One of the general form of a circle is: $$ x^2 + y^2 + ax + by + c = 0 $$ Hence we have: \begin{align*} 4 + 9 + 2a + 3b + c &= 0 \,\,\, \text{ This is for the first point. } \\ 9 + 4 + 3a + 2b + c &= 0 \,\,\, \text{ This is for the second point. }\\ 16 + 9 - 4a + 3b + c &= 0 \,\,\, \text{ This is for the third point. }\\ \end{align*} Now, I have a system of 3 linear equations. We put them in standard form. \begin{align*} 2a + 3b + c &= -13 \\ 3a + 2b + c &= -13 \\ -4a + 3b + c &= -16 \\ \end{align*} Now, we will solve for $a$ using Cramer's rule. \begin{align*} \begin{vmatrix} 2 & 3 & 1 \\ 3 & 2 & 1 \\ -4 & 3 & 1 \\ \end{vmatrix} &=2 \begin{vmatrix} 2 & 1 \\ 3 & 1 \\ \end{vmatrix} - 3 \begin{vmatrix} 3 & 1 \\ -4 & 1 \\ \end{vmatrix} + \begin{vmatrix} 3 & 2 \\ -4 & 3 \\ \end{vmatrix} \\ \begin{vmatrix} 2 & 1 \\ 3 & 1 \\ \end{vmatrix} &= 2 - 3 = -1 \\ \begin{vmatrix} 3 & 1 \\ -4 & 1 \\ \end{vmatrix} &= 3 + 4 = 7 \\ \begin{vmatrix} 3 & 2 \\ -4 & 3 \\ \end{vmatrix} &= 9 + 8 = 17 \\ \begin{vmatrix} 2 & 3 & 1 \\ 3 & 2 & 1 \\ -4 & 3 & 1 \\ \end{vmatrix} &= 2(-1) -3(7) + 17 = -2 - 21 + 17 = -6 \\ % \end{align*} \begin{align*} \begin{vmatrix} -13 & 3 & 1 \\ -13 & 2 & 1 \\ -16 & 3 & 1 \\ \end{vmatrix} &= -13 \begin{vmatrix} 2 & 1 \\ 3 & 1 \\ \end{vmatrix} - 3 \begin{vmatrix} -13 & 1 \\ -16 & 1 \\ \end{vmatrix} + \begin{vmatrix} -13 & 2 \\ -16 & 3 \\ \end{vmatrix} \\ \begin{vmatrix} 2 & 1 \\ 3 & 1 \\ \end{vmatrix} &= 2 - 3 = -1 \\ \begin{vmatrix} -13 & 1 \\ -16 & 1 \\ \end{vmatrix} &= -13 + 16 = 3 \\ \begin{vmatrix} -13 & 2 \\ -16 & 3 \\ \end{vmatrix} &= -39 + 32 = -7 \\ \begin{vmatrix} -13 & 3 & 1 \\ -13 & 2 & 1 \\ -16 & 3 & 1 \\ \end{vmatrix} &= -13( -1) - 3(3) - 7 = 13 - 9 - 7 = -3 \\ a &= \frac{-3}{-6} \\ a &= \frac{1}{2} \\ \end{align*} \begin{align*} \begin{vmatrix} 2 & -13 & 1 \\ 3 & -13 & 1 \\ -4 & -16 & 1 \\ \end{vmatrix} &= 2 \begin{vmatrix} -13 & 1 \\ -16 & 1 \\ \end{vmatrix} + 13 \begin{vmatrix} 3 & 1 \\ -4 & 1 \\ \end{vmatrix} + \begin{vmatrix} 3 & -13 \\ -4 & -16 \\ \end{vmatrix} \\ \begin{vmatrix} -13 & 1 \\ -16 & 1 \\ \end{vmatrix} &= -13 + 16 = 3 \\ \begin{vmatrix} 3 & 1 \\ -4 & 1 \\ \end{vmatrix} &= 3 + 4 = 7 \\ \begin{vmatrix} 3 & -13 \\ -4 & -16 \\ \end{vmatrix} &= -48 - 52 = 100 \\ \begin{vmatrix} 2 & -13 & 1 \\ 3 & -13 & 1 \\ -4 & -16 & 1 \\ \end{vmatrix} &= 2(3) + 13(7) - 100 = 6 + 91 - 100 \\ \begin{vmatrix} 2 & -13 & 1 \\ 3 & -13 & 1 \\ -4 & -16 & 1 \\ \end{vmatrix} &= -3 \\ b &= \frac{-3}{-6} \\ b &= \frac{1}{2} \end{align*} Now we take the first equation and solve for $c$. \begin{align*} 2a + 3b + c &= -13 \\ 2\left( \frac{1}{2}\right) + 3\left( \frac{1}{2}\right) + c &= -13 \\ 1 + \frac{3}{2} + c &= -13 \\ c &= - 13 - \frac{3}{2} - \frac{2}{2} \\ c &= -\frac{31}{2} \end{align*} Now we subsitute back into the general form of the circle we started with: \begin{align*} x^2 + y^2 + ax + by + c &= 0 \\ x^2 + y^2 + \left( \frac{1}{2} \right) x + \left( \frac{1}{2} \right)y - \frac{31}{2} &= 0 \\ 2x^2 + 2y^2 + x + y - 32 &= 0 \\ \end{align*} However, the book gets: $$ 2x^2 + 2y^2 + x + y - 23 = 0 $$ I claim that the point $(2,3)$ goes through my equation but not the book's equation. Am I right? Is the book right?

I copied the book's answer incorrectly. The book's answer is: $$ x^2 + y^2 + 2x + 2y - 23 = 0 $$

In addition, after correcting for the wrong constant (e.g $-16$ should have been -$25$), I was able to get the correct answer. Here is my correct answer.

Answer:

One of the general form of a circle is: $$ x^2 + y^2 + ax + by + c = 0 $$. Hence we have: \begin{align*} 4 + 9 + 2a + 3b + c &= 0 \,\,\, \text{ This is for the first point. } \\ 9 + 4 + 3a + 2b + c &= 0 \,\,\, \text{ This is for the second point. }\\ 16 + 9 - 4a + 3b + c &= 0 \,\,\, \text{ This is for the third point. }\\ \end{align*} Now, I have a system of 3 linear equations. We put them in standard form. \begin{align*} 2a + 3b + c &= -13 \\ 3a + 2b + c &= -13 \\ -4a + 3b + c &= -25 \\ \end{align*} Now, we will solve for $a$ using Cramer's rule. \begin{align*} \begin{vmatrix} 2 & 3 & 1 \\ 3 & 2 & 1 \\ -4 & 3 & 1 \\ \end{vmatrix} &=2 \begin{vmatrix} 2 & 1 \\ 3 & 1 \\ \end{vmatrix} - 3 \begin{vmatrix} 3 & 1 \\ -4 & 1 \\ \end{vmatrix} + \begin{vmatrix} 3 & 2 \\ -4 & 3 \\ \end{vmatrix} \\ \begin{vmatrix} 2 & 1 \\ 3 & 1 \\ \end{vmatrix} &= 2 - 3 = -1 \\ \begin{vmatrix} 3 & 1 \\ -4 & 1 \\ \end{vmatrix} &= 3 + 4 = 7 \\ \begin{vmatrix} 3 & 2 \\ -4 & 3 \\ \end{vmatrix} &= 9 + 8 = 17 \\ \begin{vmatrix} 2 & 3 & 1 \\ 3 & 2 & 1 \\ -4 & 3 & 1 \\ \end{vmatrix} &= 2(-1) -3(7) + 17 = -2 - 21 + 17 = -6 \\ \end{align*} \begin{align*} \begin{vmatrix} -13 & 3 & 1 \\ -13 & 2 & 1 \\ -25 & 3 & 1 \\ \end{vmatrix} &= -13 \begin{vmatrix} 2 & 1 \\ 3 & 1 \\ \end{vmatrix} - 3 \begin{vmatrix} -13 & 1 \\ -25 & 1 \\ \end{vmatrix} + \begin{vmatrix} -13 & 2 \\ -25 & 3 \\ \end{vmatrix} \\ \begin{vmatrix} 2 & 1 \\ 3 & 1 \\ \end{vmatrix} &= 2 - 3 = -1 \\ \begin{vmatrix} -13 & 1 \\ -25 & 1 \\ \end{vmatrix} &= -13 + 25 = 12 \\ \begin{vmatrix} -13 & 2 \\ -25 & 3 \\ \end{vmatrix} &= -39 + 50 = 11 \\ \begin{vmatrix} -13 & 3 & 1 \\ -13 & 2 & 1 \\ -25 & 3 & 1 \\ \end{vmatrix} &= -13(-1) - 3(12) + 11 = 13 - 36 + 11 = -12 \\ a &= \frac{-12}{-6} \\ a &= 2 \\ \end{align*} \begin{align*} \begin{vmatrix} 2 & -13 & 1 \\ 3 & -13 & 1 \\ -4 & -25 & 1 \\ \end{vmatrix} &= 2 % continue here \begin{vmatrix} -13 & 1 \\ -25 & 1 \\ \end{vmatrix} + 13 \begin{vmatrix} 3 & 1 \\ -4 & 1 \\ \end{vmatrix} + \begin{vmatrix} 3 & -13 \\ -4 & -25 \\ \end{vmatrix} \\ \begin{vmatrix} -13 & 1 \\ -25 & 1 \\ \end{vmatrix} &= -13 + 25 = 12 \\ \begin{vmatrix} 3 & 1 \\ -4 & 1 \\ \end{vmatrix} &= 3 + 4 = 7 \\ \begin{vmatrix} 3 & -13 \\ -4 & -25 \\ \end{vmatrix} &= -75 - 4(13) = -127 \\ \begin{vmatrix} 2 & -13 & 1 \\ 3 & -13 & 1 \\ -4 & -25 & 1 \\ \end{vmatrix} &= 2(12) + 13(7) - 127 = 24 + 91 - 127 \\ \begin{vmatrix} 2 & -13 & 1 \\ 3 & -13 & 1 \\ -4 & -16 & 1 \\ \end{vmatrix} &= -12 \\ b &= \frac{-12}{-6} \\ b &= 2 \end{align*} Now we take the first equation and solve for $c$. \begin{align*} 2a + 3b + c &= -13 \\ 2( 2) + 3( 2 ) + c &= -13 \\ 10 + c &= -13 \\ c &= -23 \end{align*} Now we substitute back into the general form of the circle we started with: \begin{align*} x^2 + y^2 + ax + by + c &= 0 \\ x^2 + y^2 + 2x + 2y - 23 &= 0 \\ \end{align*}

Answer 1

An alternate aproach.

Lets label our points

A = (2,3) B = (3,2) C = (-4,2)

The equation of the line containing AB is $x+y = 5$
Its perpendicular bisector is $x=y$

The equation of $AC$ is $y = 3$
And the perpendicular bisector is $x = -1$

The center of the circle then is $(-1,-1)$

And the square of the distance from the center to any of these points $3^2 + 4^2 = 5^2$

$(x+1)^2 + (y+1)^2 = 25$

Answer 2

Neither answer appears to be correct. The correct answer is $$(x+1)^2 + (y+1)^2 = 5^2,$$ or $$x^2 + 2x + y^2 + 2y - 23 = 0.$$ This can be easily verified via substitution of the three points.

Another way to solve for the equation of the circumcircle is to observe that if the circle passes through $(2,3)$ and $(3,2)$, the circumcenter must lie on the line $y = x$, because this line is the perpendicular bisector of the line segment joining these two points. Moreover, the side of the triangle joining $(-4,3)$ and $(2,3)$ has perpendicular bisector $x = \frac{-4+2}{2} = -1$. Therefore, the circumcenter is located at the intersection of these two lines, hence the circle's center is $(-1,-1)$ and its radius is simply the distance to any vertex: $$r = \sqrt{(3-(-1))^2 + (2-(-1))^2} = 5.$$ Hence the equation of the circle is as claimed above.

Answer 3

segment bisectors meet in the circle center

Answer 4

Your approach was fine, but you made a couple of computational errors.

The system should be $2a+3b+c=-13,\tag1$

$3a+2b+c=-13,\tag2$

$-4a+3b+c=-\color{red}{25}.\tag3$

Subtracting $(1$) from $(2)$ shows that $a-b=0$ or $a=b$.

Then we have $5a+c=-13$ and $-a+c=-25$; subtracting those yields $6a=12$ or $a=2$.

Can you take it from here?

Answer 5

We know from circle theorems that the perpendicular bisector of a chord of a circle passes through the circle's centre. Treating these $3$ points as $2$ chords and finding the equation of their perpendicular bisectors, the intersection point of the $2$ bisectors will be the centre of the circle.