for what value of c will the circle with equation x^2+6x+y^2-4y+c=0 have a radius of 4?
The center of the circle is found by comparison. It is $\displaystyle (g=-3,f=2)$
The Radius can be found from an equation of a general formula by $\displaystyle R=\sqrt{g^{2}+f^{2}-c}$
Therefore, the reverse can be found by inputting $\displaystyle R=4$ into $\displaystyle 4=\sqrt{g^{2}+f^{2}-c}$
Since we know the center, we input $\displaystyle g$ and $\displaystyle f$ into the above, and get: $\displaystyle 4=\sqrt{(-3)^{2}+2^{2}-c}$
Solving for $\displaystyle c$, we get $\displaystyle 16=9+4-c$ which gives a value for $\displaystyle c$ of $\displaystyle c=3$
Comparison with the general equation of a circle, which is $\displaystyle x^{2}+y^{2}+2gx+2fy+c=0$ where $\displaystyle (-g,-f)$ is the center of the circle, and the radius is per my previous post. When posting formulas, I think it's neater if you arranged them by powers before variables.