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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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Respuesta :

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
Corsaquix

Answer:

[tex]\text{C. }x^2+(y+3)^2=11[/tex]

Step-by-step explanation:

The equation of a circle with radius [tex]r[/tex] and center [tex](h, k)[/tex] is given by:

[tex](x-h)^2+(y-k)^2=r^2[/tex].

What we're given:

  • radius of [tex]\sqrt{11}[/tex]
  • center at [tex](0,-3)[/tex]

Substituting given values, we get:

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

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