What is the center and radius of the circle with equation x^2 + y^2 + 6x + 16y + 64 = 0?

1 Answer
May 4, 2016

center: (-3,-8)
radius: 3

Explanation:

Recall that the general equation of a circle is:

color(blue)(|bar(ul(color(white)(a/a)(x-h)^2+(y-k)^2=r^2color(white)(a/a)|)))

where:
x=x-coordinate
h=x-coordinate of circle's centre
y=y-coordinate
k=y-coordinate of circle's centre
r=radius of circle

Given the equation,

x^2+y^2+6x+16y+64=0

Subtract 64 from both sides.

x^2+y^2+6x+16y=-64

Grouping and bracketing the terms with x and y separately,

x^2+6x+y^2+16y=-64

(x^2+6x)+(y^2+16y)=-64

Complete the square within each bracket.

(x^2+6xcolor(white)(i)color(red)(+(6/2)^2))+(y^2+16ycolor(white)(i)color(red)(+(16/2)^2))=-64

(x^2+6xcolor(white)(i)color(red)(+9))+(y^2+16ycolor(white)(i)color(red)(+64))=-64

Add 9 and 64 to the right side of the equation.

(x^2+6xcolor(white)(i)color(darkorange)(+9))+(y^2+16ycolor(white)(i)color(darkorange)(+64))=-64color(white)(i)color(darkorange)(+64)color(white)(i)color(darkorange)(+9)

(x^2+6x+9)+(y^2+16y+64)=9

Rewrite each bracketed expression in factored form.

(x+3)^2+(y+8)^2=9

According to the general equation of a circle, the centre and radius can be determined by looking at the equation for the circle as,

(x-(color(red)(-3)))^2+(y-(color(blue)(-8)))^2=color(purple)3^2

ul("Reminder": (x-color(red)h)^2+(y-color(blue)k)^2=color(purple)r^2)

Since the centre is (color(red)h,color(blue)k) and the radius is color(purple)r, using the derived equation, we can determine that the centre of the circle is (color(red)(-3),color(blue)(-8)) and the radius is color(purple)3.