Chapter 7.3, Problem 54E

To determine

To calculate:Find the equation $x^2+y^2+Dx+Ey+F=0$ of the circle that passes through the points. To verify your result, use a graphing utility to plot the points and graph the circle.

  $\left(0,0\right)\left(0,6\right)\left(3,3\right)$

Answer to Problem 54E

The equation of the circle passing though $\left(0,0\right)\left(0,6\right)\left(3,3\right)$ is $x^2+y^2-6y=0$

Explanation of Solution

Given information:

The equation of the circle given is $x^2+y^2+Dx+Ey+F=0$ and it passes through the points $\left(0,0\right)\left(0,6\right)\left(3,3\right)$

Calculation:

As we do not know the coefficients we need to find them by substituting the points in the circle equation and by using elimination method we can know their values.

Substitute $\left(0,0\right)$ we get,

  $\begin{array}{l}{0}^2+0^2+D\left(0\right)+E\left(0\right)+F=0\\ F=0\end{array}$

We got one value now by substituting $\left(0,6\right)$ we get,

  $\begin{array}{l}{0}^2+6^2+D\left(0\right)+E\left(6\right)+0=0\\ 0+36+0+6E=0\\ 6E=-36\\ E=-6\end{array}$

Substitute $\left(3,3\right)$ , we get,

  $\begin{array}{l}{3}^2+3^2+D\left(3\right)+\left(-6\right)\left(3\right)+0=0\\ 9+9+3D-18=0\\ 3D=0\\ D=0\end{array}$

Now, by substituting this values we get the circle equation as

  $\begin{array}{l}{x}^2+y^2+\left(0\right)x+\left(-6\right)y+0=0\\ x^2+y^2-6y=0\end{array}$

The graph is as follows:

  

Conclusion:

The equation of the circle passing though $\left(0,0\right)\left(0,6\right)\left(3,3\right)$ is $x^2+y^2-6y=0$