find the center and the radius of the circle: x^2+y^2+10x+6y-27=0?

Answer 1

x^2+y^2+10x+6y-27=0

x^2+10x + y^2+6y = 27

x^2+10x + 25 + y^2 + 6y + 9 = 27 + 25 + 9

(x + 5)^2 + (y + 3)^2 = 61

Center: (-5, -3)

Radius: sqrt(61)

Answer 2

x^2 + y^2 + 10x + 6y - 27 = 0

rearranging

=>x^2 + 10x + y^2 + 6y = 27

complete the squares

=>x^2 + 2(5)x + 25 + y^2 + 2(3)y + 9 = 27 + 25 + 9

=>(x + 5)^2 + (y + 3)^2 = 61

so equation of circle is

=>(x + 5)^2 + (y + 3)^2 = (√61)^2

with center (-5, - 3) and radius √61 units

Answer 3

Let's put the equation in standard form. In order to do that we will need to complete the square from x and y.

x^2 + 10x + y^2 + 6y = 27
x^2 + 10x + 25 + y^2 + 6y + 9 = 27 + 34
(x + 5)^2 + (y + 3)^2 = 61

(x +5)^2 + (y + 3)^2 = (Sqrt 61)^2

Therefore, the center of the circle is at (-5,-3) and it has a radius of (Sqrt 61).

Answer 4

x^2 + 10x + y^2 + 6y = 27...grouping

x^2 + 10x + 25 + y^2 + 6y + 9 = 27 + 25 + 9...complete squares

(x + 5)^2 + (y + 3)^2 = 61...perfect trinomials

(x - h)^2 + (y - k)^2 = r^2...comparing with

where C (h,k) is the center & r the radius

C (-5,-3) &
r = sqrt61

Answer 5

Complete both squares
x^2+10x + y^2+6y = 27
x^2+10x+25 + y^2+6y+9 = 61
(x+5)^2+(y+3)^2=61
Center: (-5,-3)
Radius: sqrt(61)

Answer 6

x^2+y^2+10x+6y-27=0
x^2+10x+y^2+6y-27=0
x^2+(2*5)x+y^2+(2*3)y-27=0
x^2+(2*5)x+(5)^2-(5)^2+y^2+(2*3)y+(3)^2-(3)^2-27=0
x^2+(2*5)x+25-25+y^2+(2*3)y+9-9-27=0
x^2+(2*5)x+25+y^2+(2*3)y+9-9-27-25=0
(x+5)^2+(y+3)^2-61=0
(x+5)^2+(y+3)^2=61
(x+5)^2+(y+3)^2=(sqrt61)^2

centre of the circle is (-5,-3)
radius is sqrt 61 that is 7.81