Question 971021
Hard to show this without drawing, but here goes.

The line starts at (0,5) and has equal tangents to the circle, one in the first quadrant and one in the second quadrant.   Let's look at the first quadrant.  

The circle has radius sqrt (5), not 5.  

The line drawn tangent to the circle forms a perpendicular line with the radius.  That means we have a right triangle, with the hypotenuse the y-axis, and 5 units, the short leg the radius (sqrt 5), and the third leg, the tangent, with a length of 2 sqrt 5 by the Pythagorean theorem.

Using the distance formula. 
x^2+y^2=25
We know that the distance from the point on the circle to the end of the line is
sqrt {(x-0)^2 + (y-5)^2} using the distance formula.
But this is equal to sqrt  20.

Therefore, square both sides and x^2 +(y-5)^2= 20
x^2 +y^2 -10y+ 25=20.   But x^2 +y^2 =5
5-10y  +25 =20
-10y=-10
y=1

x=2

The other point is y=1, x=-2

The slope of one line is -2,  (1-5)/2
The slope of the other is +2

The equation of the lines are y=-2x+5; y= 2x +5