Chapter 13.3, Problem 65STP

To determine

To find: Equation for each Circle.

Answer to Problem 65STP

Equation of circle is

  ${\left(x+4\right)}^2+{\left(y+2\right)}^2=73$

Explanation of Solution

Given information:

End points of Diameter are

  $\left(-1,-10\right),\left(-7,6\right)$

Given

End points of Diameter are

  $\left(-1,-10\right),\left(-7,6\right)$

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

  $\begin{array}{l}\left(x_1,y_1\right)=\left(-1,-10\right)\\ \left(x_2,y_2\right)=\left(-7,6\right)\end{array}$

Therefore,

Center of circle is

  $\begin{array}{l}\left(h,k\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ \Rightarrow \left(h,k\right)=\left(\frac{-1-7}{2},\frac{-10+6}{2}\right)\\ \Rightarrow \left(h,k\right)=\left(\frac{-8}{2},\frac{-4}{2}\right)\\ \Rightarrow \left(h,k\right)=\left(-4,-2\right)\end{array}$

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

On substituting values

We get

  $\begin{array}{l}r=\sqrt{{\left(x_2-h\right)}^2+{\left(y_1-k\right)}^2}\\ \Rightarrow r=\sqrt{{\left(\left(-7\right)-\left(-4\right)\right)}^2+{\left(6-\left(-2\right)\right)}^2}\\ \Rightarrow r=\sqrt{{\left(-7+4\right)}^2+{\left(6+2\right)}^2}\\ \Rightarrow r=\sqrt{{\left(-3\right)}^2+8^2}\\ \Rightarrow r=\sqrt{9+64}\\ \Rightarrow r=\sqrt{73}\end{array}$

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

  $\begin{array}{l}{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\\ \Rightarrow {\left(x-\left(-4\right)\right)}^2+{\left(y-\left(-2\right)\right)}^2={\left(\sqrt{73}\right)}^2\\ \Rightarrow {\left(x+4\right)}^2+{\left(y+2\right)}^2=73\end{array}$

Therefore,

Equation of circle is

  ${\left(x+4\right)}^2+{\left(y+2\right)}^2=73$