Chapter 12.3, Problem 64SR

To determine

To find: Equation for each Circle.

Answer to Problem 64SR

Equation of circle is

  ${\left(x-6\right)}^2+{\left(y+1\right)}^2=25$

Explanation of Solution

Given information:

End points of Diameter are

  $\left(2,-4\right),\left(10,2\right)$

Given

End points of Diameter are

  $\left(2,-4\right),\left(10,2\right)$

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

  $\begin{array}{l}\left(x_1,y_1\right)=\left(2,-4\right)\\ \left(x_2,y_2\right)=\left(10,2\right)\end{array}$

Therefore,

Center of circle is

  $\begin{array}{l}\left(h,k\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ \Rightarrow \left(h,k\right)=\left(\frac{2+10}{2},\frac{-4+2}{2}\right)\\ \Rightarrow \left(h,k\right)=\left(\frac{12}{2},\frac{-2}{2}\right)\\ \Rightarrow \left(h,k\right)=\left(6,-1\right)\end{array}$

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

On substituting values

We get

  $\begin{array}{l}r=\sqrt{{\left(x_2-h\right)}^2+{\left(y_1-k\right)}^2}\\ \Rightarrow r=\sqrt{{\left(10-6\right)}^2+{\left(2-\left(-1\right)\right)}^2}\\ \Rightarrow r=\sqrt{{\left(10-6\right)}^2+{\left(2+1\right)}^2}\\ \Rightarrow r=\sqrt{4^2+3^2}\\ \Rightarrow r=\sqrt{16+9}\\ \Rightarrow r=\sqrt{25}\\ \Rightarrow r=5\end{array}$

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

  $\begin{array}{l}{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\\ \Rightarrow {\left(x-6\right)}^2+{\left(y-\left(-1\right)\right)}^2=5^2\\ \Rightarrow {\left(x-6\right)}^2+{\left(y+1\right)}^2=25\end{array}$

Therefore,

Equation of circle is

  ${\left(x-6\right)}^2+{\left(y+1\right)}^2=25$