What is the center and radius of the circle with equation x^2 + y^2 + 6x + 16y + 64 = 0?
1 Answer
center:
radius:
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
x^2+y^2+6x+16y=-64
Grouping and bracketing the terms with
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
(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