Find centre and radius of a circle
By Mia Morrison •
I have the question "Find the centre and radius of the circle whose equation is $x^2+y^2+x+3y-2=0$"
So I've worked out the centre to be $(-1/2, -3/2)$ which when I checked the solutions is correct however I got $9/2$ for the radius and in the solutions the radius should be $\frac{3\sqrt{2}}{2}$.
Could you explain how this is achieved for the radius?
$\endgroup$ 32 Answers
$\begingroup$You got $R^2$! you need $R$. So just take the square root.
$\endgroup$ $\begingroup$$x^2+y^2+x+3y-2=0 \Leftrightarrow (x+0.5)^2+(y+1.5)^2=4.5$
i suppose this is what you did. The common formula for circles is $(x-a)^2+(y-b)^2=r^2$ so you just need to take the root of $\frac{9}{2}$.
$\sqrt{\frac{9}{2}}=\frac{3}{\sqrt{2}}=3\frac{\sqrt{2}}{2}$
$\endgroup$
