Question

A circle in the xy-coordinate plane has the equation

�
2
+
�
2
+
6
�
−
4
=
0
x
2
+y
2
+6y−4=0 . If the equation of the circle in written in the form
�
2
+
(
�
+
�
)
2
=
�
x
2
+(y+k)
2
=c , where k and c are constants, what is the value of k?

Answer 1

Given the equation of the circle in the xy-coordinate plane as x^2 + y^2 + 6x - 4 = 0, we need to rewrite it in the form (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center of the circle and r is the radius.

Completing the square for x^2 + 6x, we get (x+3)^2 - 9. Thus, the equation becomes (x+3)^2 + y^2 - 9 = 4. Rearranging, we get (x+3)^2 + y^2 = 13.

Comparing with the required form, we get (x-h)^2 + (y-k)^2 = r^2, where h = -3, k = 0 and r^2 = 13. Thus, the value of k is 0.