Chapter 7.3, Problem 57E

To determine

The equation of the circle that passes through the points. The graphing utility to plot the points and graph the circle.

Answer to Problem 57E

  $x^2+y^2+6x-8y=0$ .

Explanation of Solution

Given information:

The points are $(-3,-1),(2,4),(-6,8)$ .

Calculation:

The given points are $(-3,-1),(2,4),(-6,8)$ .

Write the three points as a equation of the circle $x^2+y^2+Dx+Ey+F=0$ and simplify them:

  $\Rightarrow (-3,-1)$ is on the circle:

  $\begin{array}{}$ {(-3)}^2+{(-1)}^2+(-3)\cdot D+(-1)\cdot E+F=0$$ 9+1-3D-E+F=0$$ 3D+E-F=\mathrm{10..............}(1)$\end{array}$

  $\Rightarrow (2,4)$ is on the circle:

  $\begin{array}{l}{2}^2+4^2+2\cdot D+4\cdot E+F=0\\ 4+16+2D+4E+F=0\\ 2D+4E+F=-\mathrm{20............}(2)\end{array}$

  $\Rightarrow (-6,8)$ is on the circle:

  $\begin{array}{}$ {(-6)}^2+{(8)}^2+(-6)\cdot D+(8)\cdot E+F=0$$ 36+64-6D+8E+F=0$$ 6D-8E-F=\mathrm{100..........}(3)$\end{array}$

Add the first and second equation:

  $\begin{array}{l}\underset{\_}{\begin{array}{l}3D+E-F=10\\ 2D+4E+F=-20\end{array}}\\ 5D+5E=-10\\ D+E=-\mathrm{2.............}(1)\end{array}$

Add the second and third equation:

  $\begin{array}{l}\underset{\_}{\begin{array}{l}6D-8E-F=100\\ 2D+4E+F=-20\end{array}}\\ 8D-4E=80\\ 2D-E=\mathrm{20............}(2)\end{array}$

Add the first and second equation to get the value of $D$ :

  $\begin{array}{l}\underset{\_}{\begin{array}{l}2D-E=20\\ D+E=-2\end{array}}\\ 3D=18\\ D=6\end{array}$

Plug the value of $D$ in equation $(1)$ :

  $\begin{array}{l}D+E=-2\\ 6+E=-2\\ E=-8\end{array}$

The equation of the circle is $x^2+y^2+6x-8y=0$

  

Thus, the equation of circle is $x^2+y^2+6x-8y=0$ .