A circle has an equation of x^2 + y^2 + 2cy = 0. Find the value of c when the length of the tangent from (5, 4) to the circle is equal to one.

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

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

$x^2 + (y + c)^2 = c^2$
 

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