SOLUTION: find the center (h,k) and radius, r, of the cirlce. x^2+y^2+6x+8y+9=0

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Question 619433: find the center (h,k) and radius, r, of the cirlce. x^2+y^2+6x+8y+9=0
Found 2 solutions by ewatrrr, jim_thompson5910:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

 x^2+y^2+6x+8y+9=0

 (x+3)^2 - 9  + (y+4)^2  - 16 + 9

 %28x%2B3%29%5E2+%2B+%28y%2B4%29%5E2+=+16+=+4%5E2   C(-3,-4)  and r = 4

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+y^2+6x+8y+9=0

(x^2+6x)+(y^2+8y)+9=0

(x^2+6x+9-9)+(y^2+8y+16-16)+9=0

(x^2+6x+9)-9+(y^2+8y+16)-16+9=0

(x+3)^2-9+(y+4)^2-16+9=0

(x+3)^2+(y+4)^2-16=0

(x+3)^2+(y+4)^2=16

(x-(-3))^2+(y-(-4))^2=16

(x-(-3))^2+(y-(-4))^2 = 4^2

The equation is now in (x-h)^2+(y-k)^2=r^2 form where h = -3, k = -4 and r = 4. Note: this conic section is a circle.

So the center of this circle is (-3,-4) and it has a radius of 4.