Chapter EP, Problem 10.3.6EP

To determine

To calculate; to form the equation for circle satisfying given conditions

Answer to Problem 10.3.6EP

Solution: ${\left(x+\frac{7}{2}\right)}^2+{\left(y+\frac{5}{2}\right)}^2=146$

Explanation of Solution

Given the end points of diameter at $\left(-9,0\right)and\left(2,-5\right)$ , using Midpoint formula center $\left(h,k\right)$ can be found

$\begin{array}{l}\left(h,k\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{-9+2}{2},\frac{0-5}{2}\right)=\left(\frac{-7}{2},\frac{-5}{2}\right)\\ \end{array}$

Using distance formula to find radius

$r=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}=\sqrt{{\left(2-\left(-9\right)\right)}^2+{\left(-5-0\right)}^2}=\sqrt{{\left(2+9\right)}^2+25}=\sqrt{{\left(11\right)}^2+25}=\sqrt{121+25}=\sqrt{146}$

Radius of circle is $\sqrt{146}$ units, so $r^2=146$

Standard form of equation of a circle with center $\left(h,k\right)$ and radius $=r$ is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ ,

Substituting the values of $h,kandr$ in standard form of equation

$\begin{array}{l}{\left(x-\left(\frac{-7}{2}\right)\right)}^2+{\left(y-\left(\frac{-5}{2}\right)\right)}^2=146\\ {\left(x+\frac{7}{2}\right)}^2+{\left(y+\frac{5}{2}\right)}^2=146\end{array}$