# 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 #(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#.*