Write the equation of a circle whose diameter is 12 and whose center is at (-6,-3)

Guest Apr 6, 2021

#1**0 **

If the denominator of the circle is $12$, the radius is $6$.

If the center is at $(-6,-3)$, then the equation is $(x-(-6))^2+(y-(-3))^2=r^2$ for some radius r.

Putting these together, we get: $(x-(-6))^2+(y-(-3))^2=6^2$, so $\boxed{(x+6)^2+(y+3)^2=36}$

Putting it in non-standard form by expanding gives $x^2+12x+6^2+y^2+6y+3^2=36$, so $x^2+12x+y^2+6y=-9$. However, the actual equation for a circle is the standard one that was shown first, it is MUCH more common. But if you graph both equations, they give the same circle.

RiemannIntegralzzz Apr 6, 2021