Question

HELP DUE IN 10 MINS!

Which equation has the following: Center (0, -3) and radius: square root of 11.


A. x2 + (y - 3)2 = 11


B. x2 + (y - 3)2 = 121


C. x2 + (y + 3)2 = 11


D. x2 + (y + 3)2 = 121


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Answer 1

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

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
  `√(11)`
• center at
  `(0,-3)`

Substituting given values, we get:

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

Answer 2

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