Question

There are two tangents to the circle of radius centered at (0, - 1) that pass through the point (0,1). What are their equations?
Equation of a circle x^2+(y+1)^2=1^2

Answer 1

Clearly, the tangents have equation
y = 1±mx
for some slope m. So, taking m to be positive, the line meets the circle where
1-mx = √(1-x^2)-1
2-mx = √(1-x^2)
4-4mx+m^2x^2 = 1-x^2
(m^2+1)x^2 - 4mx + 3 = 0
To meet the circle in only one point, the discriminant must be zero, or
16m^2-12(m^2+1) = 0
4m^2 - 12 = 0
m = ±√3
so the lines are
y = 1±√3 x

see the graphs at

www.wolframalpha.com/input/?i=plot+x%5E2%2B%28y%2B1%29%5E2+%3D+1%2C+y+%3D+1-%E2%88%9A3+x%2C+y%3D1%2B%E2%88%9A3+x

Answer 2

The radius is 1.
DRAW IT
Lets look at the right triangle on the right (remember tan hits circle at right angle to radius)
the lower leg is 1 the radius
the vertical leg (Hypotenuse) is 2 (from -1 to +1)
so the other leg is sqrt 3
The tangent to the circle is therefore sin^-1(1/2) or 30 degrees above the -y axis, 60 degrees below x axis
the slope m is therefore -sqrt 3/1
y = (-sqrt 3) x + b
when x = 0, y = 1so b= 1
y = (- sqrt 3)x + 1
on the left the slope is +sqrt 3