Question

Find an equation of the circle that satisfies the stated conditions.
Center at the origin, passing through P(3, −6)

Answer 1

Try x^2 + y^2= 45

Answer 2

how did you get that?

Answer 3

The generic formula for a circle is:
\[(X-Xo)^{2} + (Y-Yo)^{2} = r^{2}\] Where (Xo,Yo) is the center and r is the radius.

In your case, you want it to be centered at the origin, so you make (Xo,Yo) = (0,0)
That leaves you with: \[(X-0)^{2} + (Y-0)^{2} = r^{2}\]
\[(X)^{2} + (Y)^{2} = r^{2}\]

Now you have to pick a radius so that it crosses the point (3,-6).

You use Pythagoras and the h^2 is your radius.
\[(X)^{2} + (Y)^{2} = (3)^{2}+(-6)^{2} = 9 + 36= 45\]

Answer 4

thanks :)

Answer 5

You're welcome. Hope that helped you understand. :)