Question

This one's bothering me..
Find equations for two lines through the origin that are tangent to the curve
(x^2) -4x+(y^2)+3=0

Answer 1

a picture is usually helpful.

complete the squares to determine that its a circle centered at (2,0) with a radius of 1

Answer 2

as is; you have congruent triangles formed

Answer 3

\[(x^2-4x+4)+(y^2-0)=1\]
\[(x-2)^2+(y-0)^2=1\]
therefore C(2,0), Radius=1

Tangent lines should be in the form of:
\[y=a*x\] \[y=(-a)*x\]
Gotta find 'a'!!
we get from the drawing that \[\sin a = 1/2\]
\[a=30º=\pi/4\]
bearing in mind that 'a' is the slope of the tangent line.
My solution is: \[y=+-(\pi*x)/\]