Finding equation of circle using 3 points

Question

I know there's multiple ways of solving a problem like this, but I still don't quite get how to solve them without wasting time solving unneeded complicated equations.

N(4,6) L(-4,0) P(0,8)

Write down the equation of the circle through N, P and L.

I've already browsed the forums regarding questions like this. I did find one solution which worked and I understood, but I feel like it takes too long and is unneededly complicated. Does anyone have a fairly simple and relatively quick solution to problems like these?

Answer 1

Fill in details: you need a point $\;(x,y)\;$ such that

$$(x-4)^2+(y-6)^2\stackrel I=(x+4)^2+y^2\stackrel{II}=x^2+(y-3)^2$$

Take the first equation I:

$$I\implies-8x+16-12y+36=8x+16\implies\color{red}{16x+12y=36}$$

and now take equality II:

$$8x+16=-6y+9\implies\color{red}{8x+6y=-7}$$

Now solve the red system of linear equations. What is point $\;(x,y)\;$ (if it exists at all)?

Answer 2

Circumcenter of triangle using 3 points A,B,C .

• A \begin{aligned}(A_{x},A_{y})\end{aligned}

• B \begin{aligned}(B_{x},B_{y})\end{aligned}

• C \begin{aligned}(C_{x},C_{y})\end{aligned}

  \begin{aligned}U_{x}&={\frac {1}{D}}\left[\left(A_{x}^{2}+A_{y}^{2}\right)\left(B_{y}-C_{y}\right)+\left(B_{x}^{2}+B_{y}^{2}\right)\left(C_{y}-A_{y}\right)+\left(C_{x}^{2}+C_{y}^{2}\right)\left(A_{y}-B_{y}\right)\right]\\U_{y}&={\frac {1}{D}}\left[\left(A_{x}^{2}+A_{y}^{2}\right)\left(C_{x}-B_{x}\right)+\left(B_{x}^{2}+B_{y}^{2}\right)\left(A_{x}-C_{x}\right)+\left(C_{x}^{2}+C_{y}^{2}\right)\left(B_{x}-A_{x}\right)\right]\end{aligned}

Radius of Circle : where a,b,c are length of triangle sides

use below formula : (x-p)^2+(y-q)^2=(radius)^2 where p and q are centre of circle .

Answer 3

The points all lie on a line, so there is no such circle.

You can see this by computing the slopes between two pairs of them and seeing that they are the same ($3/4$).

You could regard this line as a degenerate circle, in which case the answer then is $y=\frac34 x + 3$.

Answer 4

The equation of line passing the points N and L is:

$y = \frac {3}{4} x +3$

The point M(0, 3) locates on this line, so it is not possible to find a circle containing all these points. However if the coordinates of M is correct then the coordinates of all three points must satisfy following equation:

$(x-a)^2 +(y-b)^2=r^2$

Where C (a, b) is the center and r is the radius, Plugging the coordinates of three points in above equation gives you a system of three equation for a and b and r. using a standard method you can find a, b and n.

Answer 5

Given three noncolinear points $P_1=(x_1,y_1)$, $P_2=(x_2,y_2)$ and $P_3=(x_3,y_3)$, an equation of the circle through them is $$\det\begin{pmatrix}x^2+y^2 & x & y & 1 \\ x_1^2+y_1^2 & x_1 & y_1 & 1 \\ x_2^2+y_2^2 & x_2 & y_2 & 1 \\ x_3^2+y_3^2 & x_3 & y_3 & 1 \end{pmatrix} = 0.$$ If this isn’t simple enough, another way is to find the circle’s center $(h,k)$ by intersecting the perpendicular bisectors of two pairs of the points. Once you have that, it’s a simple matter of computing the (square of) the radius and plugging into the generic equation $(x-h)^2+(y-k)^2=r^2$.

This intersection of lines can be computed directly, without solving any equations: Let $\mathbf n_{12}=P_2-P_1$ and $\mathbf n_{23} = P_3-P_2$. Then, compute the cross product $$\left[\mathbf n_{12};-\frac12(P_1+P_2)\cdot\mathbf n_{12}\right] \times \left[\mathbf n_{23};-\frac12(P_2+P_3)\cdot\mathbf n_{23}\right].$$ The two terms in this cross product are the homogeneous vectors that represent the perpendicular bisectors of $P_1P_2$ and $P_2P_3$, respectively, and are closely related to the point-normal form of their equations. The result is the homogeneous coordinates of the circle’s center; divide through by the last component to convert to inhomogeneous Cartesian coordinates. (If the last component is $0$, the two lines are parallel—the three points are colinear.)

Using your updated example, we have $\mathbf n_{12}=(-8,-6)$ and $\mathbf n_{23}=(4,8)$, and the center of the circle is therefore $$[-8,-6,-(0,3)\cdot(-8,-6)]\times[4,8,-(-2,4)\cdot(4,8)] = [-8,-6,18]\times[4,8,-24] = [0,-120,-40],$$ which is the point $(0,3)$. The radius is easily found by computing the distance to the point $(0,8)$, so an equation of the circle is $x^2+(y-3)^2=25$.