Question

An equation of the circle whose center is the origin and

Answer 1

Answer:

$\fbox{\begin{minipage}{10em}Option C is correct\end{minipage}}$

Explanation:

A circle whose center is $(a, b)$ and radius $r$, has equation:

$(x - a)^{2} + (y - b)^{2} = r^{2}$

The center of circle is  given as origin $(0, 0)$, therefore:

$a = 0\\b = 0$

This circle passes $(3, 0)$, then the radius of circle is the distance $(d)$ between origin $(0, 0)$ and $(3, 0)$

$d = \sqrt{(0 - 3)^{2} + (0 - 0)^{2} } = \sqrt{3^{2} } = 3$

=> $r = d = 3$

Substitute $a$, $b$, and $r$ back into original equation:

$(x - 0)^{2} + (y - 0)^{2} = 3^{2}$

or

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

Hope this helps!

:)