find the radius and center of the circle, x^2+y^2+4x-12y+15=0

Question

needs reminder on how to solve an equation using a difference of two squares.

thanks.

 

Answer 1

Given equation x^2+y^2+4x-12y+15 = 0

Subtrac 15 to each side.

x^2+y^2+4x-12y+15-15 = 0-15

x^2+y^2+4x-12y = -15

Add 40 to each side.

x^2+y^2+4x-12y+40 = -15+40

x^2+y^2+4x-12y+4+36 = -15+40

x^2+4x+4+y^2-12y+36 = 25

(x+2)^2+(y-6)^2 = 5^2

(x-(-2))^2+(y-6)^2 = 5^2

Compare it to circle standard form (x-h)^2+(y-k)^2 = r^2

h = -2, k = 6, r = 5

Radius is 5, center is (-2,6).

Answer 2

Standard form of a circle equation is

Where center is (h ,k ) and radius of circle is r .

The equation is x ²  + y ²+ 4x -12y + 15 = 0

x ² + 4x + y ² -12y  = -15

To change the expression into a perfect square  add (half the x coefficient)² and add (half the y coefficient)²to each side of the expression.

Here x coefficient = 4. so, (half the x coefficient)² = (4/2)²= 4

Here y coefficient = -12. so, (half the y coefficient)² = (-12/2)²= 36

Add 4 and 36 to each side.

x ²  + 4x + 4 + y ² -12y  + 36 = -15 + 4 + 36

(x + 2)²  + (y - 6)²  = 25

(x - (-2))²  + (y - 6)²  = 5²

Compare it to standard form of circle

Center of circle = (-2,6)

Radius of circle = 5.