Question

How do you find the equation of a circle with center at the origin and passing through $\left(-6,-2\right)$?

Answer 1

Step-1: Radius of Circle:

It is given that the center of circle is origin that is $\left(0,0\right)$.

Also a point $\left(-6,-2\right)$ lies on boundary of circle. Then by using distance formula, radius of circle is :

$\begin{array}{rcl}r& =& \sqrt{{\left(0+6\right)}^2+{(0+2)}^2}\\ & \Rightarrow & r=\sqrt{36+4}\\ & \Rightarrow & r=\sqrt{40}\\ & \Rightarrow & r=2\sqrt{10}\\ & & \end{array}$

Step-2: Equation of Circle:

The general equation of circle is ${\left(x-h\right)}^2+{(y-k)}^2=r^2$, where $\left(h,k\right)$ is center of the circle.

So the equation of the given circle is :

$\begin{array}{rcl}{\left(x-0\right)}^2+{(y-0)}^2& =& {\left(2\sqrt{10}\right)}^2\\ & \Rightarrow & x^2+y^2=40\end{array}$

Therefore the equation of the circle with center at origin and passing through the given point $\left(-6,-2\right)$ is ${\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathbf{40}$.