Find the center and radius of the circle given by the equation x2 + y2 - 10x - 8y - 8 = 0?

Question

a. center (-5, 4); radius 6

b. center (-5, 4); radius 7

c. center (5, -4); radius 6

d. center (5, -4); radius 7

e. center (-5, -4); radius 6

f. center (5, 4); radius 6

g. center (-5, -4); radius 7

h. center (5, 4); radius 7

i. none of these

Answer 1

Group the x's and y's
x2 - 10x + y2 - 8y - 8 = 0
You want to get this in the form of the formula for a circle
Using just the x component, figure out what a must be to get (x-a)^2. -10x is the middle term of the resulting trinomial - which can only happen if a=5. You need +25 to get that to happen.
(x-5)^2
Similarly for y
(y-b)^2 -> b must be 4 and you need +16 to get that. Add 25 and 16 to both sides. Also add 8 so you can get rid of that term.
x^2 -10x +25 + y^2 - 8y +16 - 8 +8= 0 +25+16+8
(x-5)^2 +(y-4)^2 = 49
(x-5)^2 + (y-4)^2 = 7^2
center (5,4) radius 7.

Answer 2

x^2 - 10x + 25 + y^2 - 8y + 16 = 8 + 25 + 16
(x-5)^2 + (y-4)^2 = 49

center (5,4) r = 7 h.

Answer 3

center is (4,-3) Radius=sqrt(4x4+( -3x-3)+24)) =sqrt(16+9+24) =sq rt(49) =7 Vote as best answer if it helped