Chapter 13.3, Problem 66STP

To determine

To find: Equation for each Circle.

Answer to Problem 66STP

Equation of circle is

  ${\left(x-6.5\right)}^2+{\left(y+3.5\right)}^2=18.5$

Explanation of Solution

Given information:

End points of Diameter are

  $\left(9,0\right),\left(4,-7\right)$

Given

End points of Diameter are

  $\left(9,0\right),\left(4,-7\right)$

Center of the circle using endpoints of diameter is

  $\left(h,k\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Given

  $\begin{array}{l}\left(x_1,y_1\right)=\left(9,0\right)\\ \left(x_2,y_2\right)=\left(4,-7\right)\end{array}$

Therefore,

Center of circle is

  $\begin{array}{l}\left(h,k\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ \Rightarrow \left(h,k\right)=\left(\frac{9+4}{2},\frac{0-7}{2}\right)\\ \Rightarrow \left(h,k\right)=\left(\frac{13}{2},\frac{-7}{2}\right)\\ \Rightarrow \left(h,k\right)=\left(6.5,-3.5\right)\end{array}$

Radius of circle is $r=\sqrt{{\left(x_2-h\right)}^2+{\left(y_1-k\right)}^2}$

On substituting values

We get

  $\begin{array}{l}r=\sqrt{{\left(x_2-h\right)}^2+{\left(y_1-k\right)}^2}\\ \Rightarrow r=\sqrt{{\left(4-6.5\right)}^2+{\left(\left(-7\right)+3.5\right)}^2}\\ \Rightarrow r=\sqrt{{\left(-2.5\right)}^2+{\left(3.5\right)}^2}\\ \Rightarrow r=\sqrt{6.25+12.25}\\ \Rightarrow r=\sqrt{18.5}\end{array}$

Equation of Circle when center and radius are known is given by

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

On substituting values

We get

  $\begin{array}{l}{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\\ \Rightarrow {\left(x-6.5\right)}^2+{\left(y-\left(-3.5\right)\right)}^2={\left(\sqrt{18.5}\right)}^2\\ \Rightarrow {\left(x-6.5\right)}^2+{\left(y+3.5\right)}^2=18.5\end{array}$

Therefore,

Equation of circle is

  ${\left(x-6.5\right)}^2+{\left(y+3.5\right)}^2=18.5$