Question

An equation of the circle whose center is the origin and

which passes through the point (3,0) is
A (x – 3)2 + y2 = 9
B (x – 3)2 + y2 = 3
© x2 + y2 = 9
O x + y2 = 3

Answer 1

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

Step-by-step 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!

:)