Chapter 7, Problem 48RE

To determine

Find the equation of the circle.

Answer to Problem 48RE

  $\boxed{x^2+y^2-2x+2y-23=0}$

Graph is verified.

Explanation of Solution

Given information:

Find the equation of the circle.

  $x^2+y^2+Dx+Ey+F=0$

That passes through the points. Verify your result using a graphing utility.

  

Calculation:

Consider the given equation and graph,

  $x^2+y^2+Dx+Ey+F=0$

  

Circle passes through these points $\left(1,4\right),\left(-2,-5\right),\left(4,3\right)$ it is given.

Circle passes through the points $\left(1,4\right)$

  $\begin{array}{l}{x}^2+y^2+Dx+Ey+F=0\\ 1^2+4^2+D\times 1+E\times 4+F=0\\ 1+16+D+4E+F=0\\ D+4E+F=-\mathrm{17........}\left(1\right)\end{array}$

Similarly circle passes through the points $\left(-2,-5\right),\left(4,3\right)$

  $\begin{array}{l}{x}^2+y^2+Dx+Ey+F=0\\ {\left(-2\right)}^2+5^2+D\times -2+E\times -5+F=0\\ 4+25-2D-5E+F=0\\ -2D-5E+F=-\mathrm{29........}\left(2\right)\end{array}$

  $\begin{array}{l}{x}^2+y^2+Dx+Ey+F=0\\ {\left(4\right)}^2+3^2+D\times 4+E\times 3+F=0\\ 16+9+4D+3E+F=0\\ 4D+3E+F=-\mathrm{25........}\left(3\right)\end{array}$

Subtract equation $\left(2\right)$ from equation $\left(1\right)$ we get,

  $3D+9E=\mathrm{12.......}\left(4\right)$

Subtract equation $\left(1\right)$ from equation $\left(3\right)$ we get,

  $3D-E=-\mathrm{8........}\left(5\right)$

Subtract equation $\left(5\right)$ from equation $\left(4\right)$ we get,

  $\begin{array}{l}10E=20\\ E=2\end{array}$

Put $E=2$ in equation $\left(4\right)$ we get,

  $\begin{array}{l}3D+9\times 2=12\\ 3D+18=12\\ 3D=-6\\ D=-2\end{array}$

Put $E=2$ and $D=-2$ in equation $\left(2\right)$ we get,

  $\begin{array}{l}-2D-5E+F=-29\\ -2\times -2-5\times 2+F=-29\\ 4-10+F=-29\\ F=-23\end{array}$

Now substitute all calculated values in equation $x^2+y^2+Dx+Ey+F=0$

  $\boxed{x^2+y^2-2x+2y-23=0}$

Now solve for variable $y$ ,

  $\begin{array}{l}{x}^2+y^2-2x+2y-23=0\\ y^2+2y+\left(x^2-2x-23\right)=0\end{array}$

Apply quadratic formula,

  $\begin{array}{l}{x}^2+y^2-2x+2y-23=0\\ y^2+2y+\left(x^2-2x-23\right)=0\\ y=\frac{-2\pm \sqrt{4-4\left(x^2-2x-23\right)}}{2}\\ =-1\pm \sqrt{1-\left(x^2-2x-23\right)}\\ =-1\pm \sqrt{24-x^2+2x}\end{array}$

Use $TI-83$ calculator to verify the equation of circle.

Press $Y=$ button and enter the equation of circle,

Type $y=-1+\sqrt{24-x^2+2x}$ , $y=-1-\sqrt{24-x^2+2x}$

The display is shown as,

  

Press window button and the range of axis.

  

Now press the graph button.

  

Press 2^nd and press TRACE button and choose1 value press enter.

  

Type $X=1$ and press enter,

  

Type $X=-2$ and press enter,

  

Type $X=4$ and press enter,

  

The points lie on the circle hence verified.

Hence, the equation of circle is $\boxed{x^2+y^2-2x+2y-23=0}$