SOLUTION: Find the center and the radius of the circle whose equation is given: x^2+y^2+6x+8=0 Center = (_,_) Radius = _

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Question 55199: Find the center and the radius of the circle whose equation is given: x^2+y^2+6x+8=0
Center = (_,_) Radius = _
Found 3 solutions by Earlsdon, venugopalramana, stanbon:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
The general form of the equation for a circle with its center at (h, k) and radius of r is:
%28x-h%29%5E2%2B%28y-k%29%5E2+=+r%5E2
Let's get your equation into this form, then we can find the center (h, k) and the radius (r). Starting with the given equation:
x%5E2%2By%5E2%2B6x%2B8+=+0 Complete the square in the x-terms by adding 1 to the 8 and 1 to the other side of the equation:
%28x%5E2%2B6x%2B9%29+%2B+y%5E2+=+1 Factor the parentheses.
%28x%2B3%29%5E2+%2B+y%5E2+=+1 Compare with the general form:
%28x-h%29%5E2+%2B+%28y-k%29%5E2+=+r%5E2
You can see that the center (h, k) is (-3, 0) because h = -3 and k = 0 and the radius (r) is 1 because r%5E2+=+1 and r+=+sqrt%281%29.

Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
Find the center and the radius of the circle whose equation is given: x^2+y^2+6x+8=0
[X^2+2(X)(3)+3^2]-3^2+Y^2+8=0
(X+3)^2+Y^2=1
COMPARING WITH STD EQN.
(X-H)^2+(Y-K)^2 =R^2
CENTRE = (H,K) = (-3,0)
RADIUS =R =1

Center = (_,_) Radius = _

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find the center and the radius of the circle whose equation is given: x^2+y^2+6x+8=0
Rewrite as follows:
x^2+6x+?+y^2 = -8+?
Complete the square to get:
x^2+6x+9 + y^2 = -8+9
Factor to get:
(x+3)^2 + (y-0)^2 = 1
Center at (-3,0)
Radius = 1
Cheers,
Stan H.