$x^2 + y^2 + 2cy = 0$

$x^2 + (y^2 + 2cy + \color{#FC6255}{c^2}) = \color{#FC6255}{c^2}$

$x^2 + (y + c)^2 = c^2$

Note:

The *standard equation* of circle with center at (*h*, *k*) and length of radius equal to *r* is given by $(x - h)^2 + (y - k)^2 = r^2$.

Hence,

Center is at (0, -*c*), and radius *r* = *c*.

From the figure, the distance between (0, -*c*) and (5, 4) is

$d^2 = (5 - 0)^2 + (4 + c)^2$

$d^2 = 25 + (4 + c)^2$

And by Pythagorean Theorem for the right triangle *CTP,*

$c^2 + 1^2 = d^2$

$c^2 + 1 = 25 + (4 + c)^2$

$c^2 + 1 = 25 + (16 + + 8c + c^2)$

$8c = -40$

$c = -5$ ← *answer*