well, something about analytical geometry?

Question

What is the radius of a circle with equation x^2+y^2-4x+2y-20=0
well, I got the answer of 5
then, the second qiz is, What is the centeral of the circle above?
I got the answer of(-2,1) but the answer key shows it supposes to be (2,-1) please tell me why? did I have some error in my formula?

Answer 1

x² + y² - 4x + 2y - 20 = 0
(x² - 4x + 4) + (y² + 2y + 1) = 20 + 4 + 1
(x - 2)² + (y + 1)² = 25

The circle has center (2,-1) and radius 5.

Answer 2

Yeah, you got the signs backwards ☺ Re-write it as
x² - 4x + y² + 2y = 20 Then
x² - 4x + 4 + y² + 2y + 1 = 25 so the x and y terms are perfect squares. then it's
(x-2)² + (x + 1)² = 5² (which you seem to have gotten)
But the equation for a circle with center at (h,k) is
(x - h)² + (y-k)² = r² so the center of this circle would be at (2,-1).

But you got the idea. Hell, *I* drop signs all the time. Just practice a bit more and you'll get good at (I mean solving problems, not dropping signs ☺)


Doug

Answer 3

If you have the equation

x^2+y^2-2ax-2by+c=0 The center is (a,b) and r^2=a^2+b^2-c

In your case -2a=-4 so a=2 and -2b=2 so b=-1 and r^2=4+1+20
s0r=5