A circle in the xy-coordinate plane has the equation x2 + y2 + 6y - 4 = 0. If the equation of the circle is written in the form x2 + (y+k)2 = c, where k and c

Question

A circle in the xy-coordinate plane has the equation x2 + y2 + 6y - 4 = 0. If the equation of the circle is written in the form x2 + (y+k)2 = c, where k and c are constants, what is the value of k?

 

A circle in the xy-coordinate plane has the equation x2 + y2 + 6y - 4 = 0. If the equation of the circle is written in the form x2 + (y+k)2 = c, where k and c are constants, what is the value of k?

 

K= ?

Answer 1

x²+y²+6y-4=0

x²+(y²+6y)=4

complete the square with y²+6y

(1/2)(6)=3 and 3²=9

add 9 to both sides

x²+(y²+6y+9)=4+9

x²+(y+3)²=13

the value of k is 3