Question

Given the equation (x+2)^2 + (y+3)^2 =4 , (a) find the center and radius; (b) graph; (c) find the intercepts, if any. Please show all of your work.

Answer 1

a. Center = (h,k) = (-2,-3); Radius = r = sqrt[4] = 2

Answer 2

Thank you. Can you show me how you got this answer?

Answer 3

The equation of a circle is x^2+y^2=r^2 or (x-h)^2 + (y-k)^2 = r^2. So, for (x+2)^2 + (y+3)^2 =4, you take the opposite of what's being manipulated in the parentheses (for (x+2) = (x-h), so in this case 2 = -h, so -2 = h... if that makes sense.) Same with the other one. (y+3) = (y-k), so 3 = -k, so -3 = k.

Answer 4

This is great! Thank you very much.

Answer 5

In other words, think of what would make that parentheses equal 0. For x: (x+2), -2 would make that parentheses equal 0. For the radius, 4 = r^2, so sqrt[4] = r = 2.

Answer 6

And for C., the only intercept would be (0,-3) as shown on the graph.

Answer 7

Ok thank you