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

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

amistre64 · Answer

a picture is usually helpful.

complete the squares to determine that its a circle centered at (2,0) with a radius of 1
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amistre64 · Answer

as is; you have congruent triangles formed

amistre64 · Answer

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anonymous · Answer

$$(x^2-4x+4)+(y^2-0)=1$$
$$(x-2)^2+(y-0)^2=1$$
therefore C(2,0), Radius=1
|dw:1333226481119:dw|
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: $$y=+-(\pi*x)/$$