Chapter B, Problem 38E

To determine

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

Answer to Problem 38E

Equation of circle is $x^2+y^2+2x-10y-104=0$

Explanation of Solution

Given information: Centre $\left(-1,5\right)$ and passes through $\left(-4,-6\right)$

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:

  $Center,\left(h,k\right)=\left(-1,5\right)$

Equation of circle is given as

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

Substituting the values in above equation:

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

Also, given that circle passes through the point $\left(-4,-6\right)$, thus the point must satisfy equation (1),

  $\begin{array}{l}{\left(-4+1\right)}^2+{\left(-6-5\right)}^2=r^2\\ r^2={\left(-3\right)}^2+{\left(-11\right)}^2\\ r^2=9+121\\ r^2=130\end{array}$

Substituting the value of $r^2$ in equation (1) to get the equation of circle

  $\begin{array}{l}{\left(x+1\right)}^2+{\left(y-5\right)}^2=130\\ x^2+1+2x+y^2+25-10y=130\\ x^2+y^2+2x-10y+\left(1+25-130\right)=0\\ x^2+y^2+2x-10y-104=0\end{array}$

Conclusion:

Hence, equation of circle is $x^2+y^2+2x-10y-104=0$