Chapter 12, Problem 5CR

To determine

To find: The standard equation of the circle.

Answer to Problem 5CR

The equation of the circle is $x^2+y^2+2x-4y=20$ .

Explanation of Solution

Given information:

In given problem, centre of the circle is $\left(-1,2\right),\left(3,5\right)$ .

Calculation:

The general equation of circle is,

  ${\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2=r$ …… $\left(1\right)$

Where,

  $\left(x_1,y_1\right)$ Is centre of circle and $r$ is radius of circle.

In given problem, centre of the circle is $\left(-1,2\right),\left(3,5\right)$ .

The radius of circle can be calculated as follows.

  $\begin{array}{c}r=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left[3-\left(-1\right)\right]}^2+{\left[5-2\right]}^2}\\ =\sqrt{{\left(3+1\right)}^2+3^2}\\ =\sqrt{4^2+3^2}\end{array}$

  $\begin{array}{c}r=\sqrt{16+9}\\ =\sqrt{25}\\ =5\end{array}$

If we put these values in equation $\left(1\right),$ we get the equation of the line as follows.

  $\begin{array}{c}{\left[x-\left(-1\right)\right]}^2+{\left[y-2\right]}^2=5^2\\ {\left(x+1\right)}^2+{\left(y-2\right)}^2=25\\ x^2+2x+1+y^2-4y+4=25\\ {\left(x+1\right)}^2+{\left(y-2\right)}^2=25\end{array}$

  $\begin{array}{c}{x}^2+2x+1+y^2-4y+4=25\\ x^2+y^2+2x-4y=25-5\\ x^2+y^2+2x-4y=20\end{array}$

Therefore, the equation of the circle is $x^2+y^2+2x-4y=20$ .