Question

find the general form of the equation of the circle.
Center at the origin and containing the point (-2,3)

Answer 1

Generally, the formula for any circle centred on the point \((a,b)\) is:
\[(x-a)^2+(y-b)^2=r^2\]
So if it is centred on the origin, you have
\[x^2+y^2=r^2\]
And if you know a point, you can find out the radius by plugging in the x and y co-ordinates to find your radius and therefore complete the equation:
\[(-2)^2+(3)^2=r^2\]
\[4+9=r^2\]
\[13=r^2\]
\[r=\sqrt{13}\]
So therefore, the formula is
\[x^2+y^2=13\]

Answer 2

thanks!!!!!

Answer 3

Anytime man. You can also solve for x by doing:

\[x^2+y^2=13\]
\[y^2=13-x^2\]
\[y=\pm\sqrt{13-x^2}\]