Question

give the equation of the circle centered at the origin and passing through the point (-3, 0).

Explanation:

Step1: Recall circle equation

The standard equation of a circle centered at the origin $(0,0)$ is $x^{2}+y^{2}=r^{2}$, where $r$ is the radius.

Step2: Calculate the radius

The circle passes through $(-3,0)$. The distance from the origin $(0,0)$ to the point $(-3,0)$ is the radius. Using the distance formula $d = \sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}$, here $x_1 = 0,y_1 = 0,x_2=-3,y_2 = 0$. So $r=\sqrt{(-3 - 0)^{2}+(0 - 0)^{2}}=\sqrt{(-3)^{2}} = 3$.

Step3: Write the circle equation

Substitute $r = 3$ into $x^{2}+y^{2}=r^{2}$, we get $x^{2}+y^{2}=9$.

Answer:

$x^{2}+y^{2}=9$