+0

# coordinates

+1
108
1

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

Apr 6, 2021

#1
+484
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.

Apr 6, 2021