Question

Circle P has a radius of 8 units with center P at (2, -1). Which equation defines circle P?

(x + 2)2 + (y – 1)2 = 64

(x – 2)2 + (y + 1)2 = 64

(x – 2)2 + (y + 1)2 = 8

(x + 2)2 + (y – 1)2 = 8

Answer 1

please helpppppp!!!!! I'm soooo stuck on how to solve and find equations!

Answer 2

Use the fact that circles are loci of points equidistant from their center, and Pythagoras' theorem to calculate said distance.

Answer 3

so you mean I would use this theorem? a^2 + b^2= c^2

Answer 4

I don't really undestand how, can you please lead me through it @Cheese3.1416

Answer 5

The general equation of any circle is

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

where (h,k) is the center and r is the radius

Answer 6

In your case, the given center is (2, -1)

so (h,k) = (2, -1) which means h = 2 and k = -1

Answer 7

You're also given a radius of 8, so r = 8

Answer 8

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

(x-2)^2 + (y-(-1))^2 = r^2 ... plug in h = 2 and k = -1

(x-2)^2 + (y+1)^2 = r^2 ... simplify

(x-2)^2 + (y+1)^2 = 8^2 ... plug in r = 8

(x-2)^2 + (y+1)^2 = 64 ... square 8 to get 64

Answer 9

ohhhhh, ok that's where the pythagoreans theorem comes into play; that's the basis correct?
Then you input the information in said places... I see.

Answer 10

So the equation of the circle with center (2, -1) with a radius of 8 is

(x-2)^2 + (y+1)^2 = 64

Answer 11

yes basically you can think of any point on the circle as a point on a right triangle like this

Answer 12

since a^2 + b^2 = c^2 by the pythagorean theorem, we can say this



which leads us to x^2 + y^2 = r^2