Question

Find the standard form of the equation of each circle.

Center at the origin and containing the point (-2, 3)

Short answer

The standard form of the equation of a circle center at the origin and containing point (-2,3) is$x^2+y^2=13$

Step-by-step solution

Step 1. Given information

Here is a circle with center at the origin and containing the point (-2,3) is given.

we have to form the standard form of the equation of a circle

Step 2. Find radius r of the circle

To find the equation of a circle we need to know the radius r. Since point (-2, 3) is given., the radius is equal to the distance from (-2, 3) to the center (0,0). So radius

$$\begin{gathered} r=\sqrt{{(-2-0)}^2+{(3-0)}^2}\sin cedis\tan cebetween2pointsfrom\left(x_{1,}{y}_1\right)to\left(x_{2,}{y}_2\right)is\sqrt{{\left(x_{2-}{x}_1\right)}^2+{\left(y_{2-}{y}_1\right)}^2} \\ =\sqrt{4+9}here\left(x_{1,}{y}_1\right)is(0,0)and\left(x_{2,}y2\right)is(-2,3). \\ =\pm \sqrt{13} \\ Henceradiusis\pm \sqrt{13.} \end{gathered}$$

Step 3. Formation of the standard form of the equation  of  a circle 

The standard form of the equation of a circle is of radius r with center at the

origin (0,0) is

$x^2+y^2=r^2$.

Here radius r is $\sqrt{13}$, thus the standard form of the equation of circle islocalid="1646842721588" $$\begin{gathered} x^2+y^2={\left(\sqrt{13}\right)}^2 \\ {x}^2+y^2=13 \end{gathered}$$