Question

find the center and the radius of x^2+y^2+6x-16y+48=0

Answer 1

Do you know how to complete the square?

Answer 2

um is that when you factor it perfectly?

Answer 3

\(x^2+y^2+6x-16y+48=0\\x^2+6x+y^2-16y+48=0\\(x+3)^2-9+y^2-16y+48=0\)

Answer 4

does this make sense?

Answer 5

\(x^2+bx = (x+\frac{b}{2})^2-(\frac{b}{2})^2\)

Answer 6

so

\(x^2+6x = (x+3)^2-9\)

Answer 7

similarly

\(y^2-16y = (y-8)^2-64\)

Answer 8

yes so far it is

Answer 9

so

\(x^2+6x+y^2-16y+48=0\\(x+3)^2-9+(y-8)^2-64+48=0\\(x+3)^2+(y-8)^2-25=0\\(x-(-3))^2+(y-8)^2=25\\(x-(-3))^2+(y-8)^2=5^2\)

Answer 10

the equation of a circle with center \((a,b)\) and radius \(r\) is \((x-a)^2+(y-b)^2=r^2\)

Answer 11

so is the center is (-3,8) ?

Answer 12

correct

Answer 13

and the radius is just 5?

Answer 14

correct