Question

Find the circle that runs through the points (5,5),(4,6),and (6,2). Write your equation in the form$\mathit{a}\mathbf{+}\mathit{b}\mathit{x}\mathbf{+}\mathit{c}\mathit{y}\mathbf{+}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathbf{0}$. Find the centre and radius of this circle.

Short answer

The required equation is $x^2+y^2-2x-4y=20$, which is the circle with centre (1,2) and radius 5 that runs through the points (5,5),(4,6) and (6,2).

Step-by-step solution

Consider the points and substitute them in the equation

The centre and the radius of the circle of form $\mathbf{(}\mathit{x}\mathbf{-}\mathit{a}{\mathbf{)}}^{\mathbf{2}}\mathbf{+}\mathbf{(}\mathit{y}\mathbf{-}\mathit{b}{\mathbf{)}}^{\mathbf{2}}\mathbf{=}{\mathit{r}}^{\mathbf{2}}$is $\left(a,b\right)$and rrespectively.

Substitute the given points in the equation$a+bx+cy+x^2+y^2=0$

$$\begin{gathered} 5^2+5^2+a+\left(5\right)b+\left(5\right)c=0 \\ a+5b+5c=-50 \end{gathered}$$

$$\begin{gathered} 4^2+6^2+a+\left(4\right)b+\left(6\right)c=0 \\ a+4b+6c=-52 \end{gathered}$$

$$\begin{gathered} 6^2+2^2+a+\left(6\right)b+\left(2\right)c=0 \\ a+6b+2c=-40 \end{gathered}$$

Rearrange the terms of the above equations

Consider the simplified equations.

$$\begin{gathered} -50=a+5b+5c......\left(1\right) \\ -52=a+4b+6c......\left(2\right) \\ -40=a+6b+2c......\left(3\right) \end{gathered}$$

Solve the above equations (1), (2) and (3)

Arrange the equations (1), (2) and (3) into matrix form:

$\left(\begin{array}{ccc}1& 5& 5\\ 1& 4& 6\\ (1& 6& 2\end{array}\right)\left(\begin{array}{c}a\\ b\\ c\end{array}\right)=\left(\begin{array}{c}-50\\ -52\\ -40\end{array}\right)$

Upon solving the values of$a,b$ and are obtained as$a=-20,b=-2,c=-4$

Substitute these values in the given equation of the circle.

$$\begin{gathered} x^2+y^2+\left(-20\right)+\left(-2\right)x+\left(-4\right)y=0 \\ {x}^2+y^2-20-2x-4y=0 \end{gathered}$$

Rearrange the terms of the above equation

Rearrange the resulting equation into standard form of circle as:

$$\begin{gathered} x^2+y^2-20-2x-4y=0 \\ {x}^2+y^2-2x-4y=20 \\ (x^2+1-2x)+(y^2+4-4y)=20+1+4 \\ (x-1{)}^2+(y-2{)}^2=25 \end{gathered}$$

Compare the above equation with the standard equation of the circle.

$$\begin{gathered} {\left(x-a\right)}^2+{\left(y-b\right)}^2=r^2\iff {\left(x-1\right)}^2+{\left(y-2\right)}^2=5^2 \\ \left(a,b\right)=\left(1,2\right) \\ r=5 \end{gathered}$$

The required equation is $x^2+y^2-2x-4y=20$, which is the circle with centre (1,2) and radius 5 that runs through the points (5,5),(4,6) and (6,2).