i need the radius and center of a circle when the equation is x^2+y^2+4x-8y-105=0?

Question

can somebody help me please

Answer 1

x^2+y^2+4x-8y-105=0
x^2+4x+4+y^2-8y+16-125=0
(x+2)^2 + (y-4)^2 = 125
Center (-2,4)
Radius = sqrt(125) = 5sqrt(5)

Answer 2

x^2+y^2+4x-8y-105=0
rewriting the equation
x^2 + 4x + y^2 - 8y = 105
performing completing the square to have a square of a binomial, we have

(x^2 + 4x + 4) + (y^2 - 8y + 16) = 105 + 4 + 16
(because we added 4 and 16 to the left side, we do the same for the write side so that the two sides remain equal)
Factoring the perfect square trinomials on the left side, we have

(x + 2)^2 + (y -4)^2 = 125

The standard form of a circle with center at (h,k) is
(x-h)^2 + (y-k)^2 = r^2

Therefore, from the equation we obtained, h = -2 , k = 4 , and
r^2 = 125
r = √125=√(25*2)=√25*√5=5√5

so the center is at (-2,4) and the radius is 5√5

Answer 3

standard equation of circle = (x-h)^2+(y-k)^2 = r^2---eqn(1)

where center = (h,k) and r = radius

x^2 + y^2 + 4x - 8y - 105 = 0

x^2 + 4x +y^2 - 8y - 105 = 0

x^2 + 4x + 4 - 4 + y^2 - 8y + 16 - 16 - 105 = 0

(x+2)^2 - 4 + (y -4)^2 -16 - 105 = 0

=>(x+2)^2 + (y -4)^2 = 125

=>(x+2)^2 + (y -4)^2 = [5sqrt(5)]^2

comparing withy standard equation(1) of circle

center = (-2, 4) and radius = 5 sqrt(5)

Answer 4

use completing sq

x^2 + 4x + 4 = (x+2)^2
y^2 - 8y +16 = (y-4)^2

eq becomes

(x+2)^2 + (y-4)^2 = 125
center of circle = (-2,4)
radius = sqrt(125) = 5sq(5) = 11.18