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