Chapter 1.4, Problem 35AYU

To determine

To find: the standard form of the equation of the circle

Answer to Problem 35AYU

The required circle equation is $x^2+y^2=13$

Explanation of Solution

Given:

Center point: $\left(-2,3\right)$

Formula Used:

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

Calculation:

The objective of this problem is to determine the standard form of the equation of the circle with center at the origin and containing the point $\left(-2,3\right)$ .

The general equation of a circle with center at $\left(h,k\right)$ and radius $r$ is,

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

In this problem, the center is at the origin $\left(0,0\right)$ , so the equation of a circle will be:

  $\begin{array}{l}{\left(x-0\right)}^2+{\left(y-0\right)}^2=r^2\\ x^2+y^2+r^2\mathrm{...........}\left(1\right)\end{array}$

Itis given that circle contains the point $\left(-2,3\right)$ , so plugin $x=-2$ , and $y=3$ in the equation $\left(1\right)$ .

  $\begin{array}{l}{\left(-2\right)}^2+{\left(3\right)}^2=r^2\\ 4+9=r^2\\ r^2=13\end{array}$

Now, substitute the value $r^2=13$ in the equation $\left(1\right)$ .

  $x^2+y^2=13$

Therefore, the required circle equation is, $x^2+y^2=13$

Conclusion:The required circle equation is $x^2+y^2=13$