Q.35 - Precalculus Enhanced with Graphing Utilities

Question

Find the standard form of the equation of each circle.

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

Step-by-Step Solution

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 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 

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

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 is $$\begin{gathered} x^2+y^2={\left(\sqrt{13}\right)}^2 \\ {x}^2+y^2=13 \end{gathered}$$