Chapter B, Problem 37E

To determine

To calculate: To find the equation of circle that satisfies the given condition

Answer to Problem 37E

Equation of circle is $x^2+y^2-6x+2y-15=0$

Explanation of Solution

Given information: Centre $\left(3,-1\right)$ and radius $5$

Formula Used:

Equation of circle with center as $\left(h,k\right)$ and radius $r$ is given as

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

Calculation:

Given the below details:

  $\begin{array}{l}Center,\left(h,k\right)=\left(3,-1\right)\\ Radius,r=5\end{array}$

Equation of circle is given as

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

Substituting the values in above equation:

  $\begin{array}{l}{\left(x-3\right)}^2+{\left(y-\left(-1\right)\right)}^2=5^2\\ {\left(x-3\right)}^2+{\left(y+1\right)}^2=5^2\\ x^2+9-6x+y^2+1+2y=25\\ x^2+y^2-6x+2y+\left(9+1-25\right)=0\\ x^2+y^2-6x+2y-15=0\end{array}$

Conclusion:

Hence, equation of circle is $x^2+y^2-6x+2y-15=0$