Chapter 7, Problem 46RE

To determine

To calculate: The equation of the circle is given as $x^2+y^2+Dx+Ey+F=0$ which goes through the given points and plot the points and graph the circle.

Answer to Problem 46RE

The equation of the circle is $x^2+y^2-4x+4y-1=0$

The graph is

  

Explanation of Solution

Given information:

The equation of the circle is given as $x^2+y^2+Dx+Ey+F=0$ and it passes through the points $\left(2,1\right),\left(-1,-2\right),\left(5,-2\right)$ from the figure.

  

Calculation:

First let us substitute $\left(2,1\right)$ in the given equation.

  $\begin{array}{l}{\left(2\right)}^2+{\left(1\right)}^2+D\left(2\right)+E\left(1\right)+F=0\\ 4+1+2D+E+F=0\\ 2D+E+F+5=0\cdots \left(i\right)\end{array}$

Let us substitute $\left(-1,-2\right)$ to get other equation.

  $\begin{array}{l}{\left(-1\right)}^2+{\left(-2\right)}^2+\left(-1\right)D+\left(-2\right)E+F=0\\ 1+4-D-2E+F=0\\ -D-2E+F+5=0\cdots \left(ii\right)\end{array}$

Let us substitute $\left(5,-2\right)$ for next equation

  $\begin{array}{l}{\left(5\right)}^2+{\left(-2\right)}^2+\left(5\right)D+\left(-2\right)E+F=0\\ 25+4+5D-2E+F=0\\ 5D-2E+F+29=0\cdots \left(iii\right)\end{array}$

Now subtract equations $\left(i\right)$ and $\left(ii\right)$ , we get,

  $\begin{array}{l}2D+E+F+5+D+2E-F-5=0\\ 3D+3E=0\\ D+E=0\cdots \left(iv\right)\end{array}$

Now subtract equations $\left(ii\right)$ and $\left(iii\right)$ , we get

  $\begin{array}{l}-D-2E+F+5-5D+2E-F-29=0\\ -6D-24=0\\ D=-4\end{array}$

Substitute the value $D=-4$ in $D=-4\left(iv\right)$ to get the value of E

  $\begin{array}{l}D+E=0\\ -4+E=0\\ E=4\end{array}$

Put $D=-4$ and $E=4$ in $\left(iii\right)$ , we get

  $\begin{array}{l}-D-2E+F+5=0\\ -\left(-4\right)-2\left(4\right)+F+5=0\\ 4-8+F+5=0\\ F=-1\end{array}$

Hence the equation of the circle is $x^2+y^2-4x+4y-1=0$

The graph is as follows

  

Conclusion:

The equation of the circle is $x^2+y^2-4x+4y-1=0$