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Question 2 (Worth 5 points) (07.03 LC) A circle is centered at (-3, 2) and has a radius of 2. Which of the following is the equation for this circle? $(x+3)^2+(y-2)^2 = 2$ $(x-3)^2+(y+2)^2 = 4$ $(x+3)^2+(y-2)^2 = 4$ $(x-3)^2 + (y+2)^2 = 2$ Points earned on this question: 0

          Question 2 (Worth 5 points)
(07.03 LC)
A circle is centered at (-3, 2) and has a radius of 2. Which of the following is the equation for this circle?
$(x+3)^2+(y-2)^2 = 2$
$(x-3)^2+(y+2)^2 = 4$
$(x+3)^2+(y-2)^2 = 4$
$(x-3)^2 + (y+2)^2 = 2$
Points earned on this question: 0

Question 2 (Worth 5 points)
(07.03 LC)
A circle is centered at (-3, 2) and has a radius of 2. Which of the following is the equation for this circle?
(x+3)^2+(y-2)^2 = 2
(x-3)^2+(y+2)^2 = 4
(x+3)^2+(y-2)^2 = 4
(x-3)^2 + (y+2)^2 = 2
Points earned on this question: 0

Added by Richard R.

Geometry A Common Core Curriculum

Ron Larson, Laurie Boswell 1st Edition

Instant Answer

Solved by Expert Carrie Godfrey

Step 1

Plugging these values into the equation, we get: (x + 3)^2 + (y - 2)^2 = 2^2 Simplifying further, we have: (x + 3)^2 + (y - 2)^2 = 4 Show more…

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'it was wrong and i need help finding the right answer Question 2 (Worth 5 points) (07.03 LC) Acircle is centered at (-3,2) and has a radius of 2. Which of the following is the equation for this circle? 0 (x + 3)2 + (y - 2)2 = 2 (x-392+ (y+2)2=4 0 (x+3)2 + (y ~ 2)2 = 4 O (x - 3)2 + (y + 2)2 = 2 Points earned on this question: 0'