what is the radius of X2+Y2- 8X + 6Y +9 =0?

First rearrange: x2 - 8x + y2 + 6y = -9

Then complete the squares

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

16 will be the radius squared, so the radius is 4 (and the center is (4, -3)).

Step a million: collect x and y words separately. Step 2: comprehensive the sq.. x^2 + y^2 - 8x + 6y - 11 = 0 x^2 - 8x + y^2 + 6y - 11 = 0 (x^2 - 8x + sixteen - sixteen) + (y^2 + 6y + 9 - 9) - 11 = 0 (x^2 - 8x + sixteen) - sixteen + (y^2 + 6y + 9) - 9 - 11 = 0 (x - 4)^2 - sixteen + (y + 3)^2 - 9 - 11 = 0 (x - 4)^2 + (y + 3)^2 - 36 = 0 (x - 4)^2 + (y + 3)^2 = 36 So r^2 = 36. Then r = 6. B

Okay, (8/2)^2 + (6/2)^2 - 9 needs to be added to both sides.

So, that gives you x2+y2-8x+6y+9+16+9-9=16+9-9.

And rearranging, x2-8x+16+y2+6y+9 = 16.

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

So, the radius is 4.

X^2+Y^2- 8X + 6Y +9 =0

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

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

center at (4, -3) radius=4

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

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

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

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

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

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

since,general eq of circle is

(x-rx)^2+(y-ry)=(r)^2

where rx=x coordinate of circle &ry=y coordinate of circle

and r=radius

radius=4

it's coordinates r=(4,-3)