Question

The equation of Circle P is ( x − 3)2 + (y + 2)2 = 34 and the equation of circle Q is (x − 4)2 + (y + 5)2 = 52. What is the distance between the centers of these circles? The equation of Circle P is ( x − 3)2 + (y + 2)2 = 34 and the equation of circle Q is (x − 4)2 + (y + 5)2 = 52. What is the distance between the centers of these circles? @Mathematics

Answer 1

Center of Circle P is (3,-2) and the Center of Circle Q is (4,-5) And the distance formula is \[\sqrt{(x2 - x1)^2 + (y2 - y1)^2}\] Just plug in the numbers and simplify.

Answer 2

Wow thanks

Answer 3

ive seen this somewhere before so i should be able to get it

Answer 4

Its pretty simple. Do you know how i got the centers of the circles?

Answer 5

no how?

Answer 6

Ok, the standard equation of a circle is \[(x-h)^2 + (y-k)^2 = r^2\] Where the center is at \[(h,k)\]

Answer 7

(0,0) ?

Answer 8

So your equation looks like this \[(x-3)^2 + (y+2)^2 = 34\] So comparing this with the standard equation of a circle, what would h and k be?

Answer 9

i cant figure it out

Answer 10

\[(x-h)^2 + (y-k)^2 = r^2 = standard\]\[(x-3)^2 + (y+2)^2 = 34 = yours\]

Answer 11

They are exactly the same format.. all you do is find h in the standard and then follow it down to your problem.

Answer 12

so h would be 3 and k would be -2

Answer 13

oh ok

Answer 14

the tricky part is the -2.. since the standard equation reads \[(y-k)^2\] but you have \[(y+2)^2\] so in order to get that you subtract a negative like this \[(y-(-2))^2\] subtracting a negative is like adding so you end up with \[(y+2)^2\]

Answer 15

yeah its like multiplaing a negative by a negative in a sense so it remains positive.

Answer 16

multiplying i mean

Answer 17

right

Answer 18

So you think you understand how to find the center?

Answer 19

i think so

Answer 20

thanks for the help