Question

HELP DUE IN 10 MINS!

Answer 1

D. I believe bc part of the equation is (y-y one) and y one is -3 and two negatives make a positive. And 11 squared it 121

Answer 2

Answer:

$\text{C. }x^2+(y+3)^2=11$

Step-by-step explanation:

The equation of a circle with radius $r$ and center $(h, k)$ is given by:

$(x-h)^2+(y-k)^2=r^2$.

What we're given:

• radius of $\sqrt{11}$
• center at $(0,-3)$

Substituting given values, we get:

$(x-0)^2+(y-(-3))^2=\sqrt{11}^2,\\\boxed{\text{C. }x^2+(y+3)^2=11}$