SOLUTION: The line y= mx + c is tangent to the curve x^2 + y^2 = 4. Prove that 4m^2 = c^2

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Question 1209461: The line y= mx + c is tangent to the curve x^2 + y^2 = 4. Prove that 4m^2 = c^2 - 4
Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!

x^2 + y^2 = 4
x^2 + (mx+c)^2 = 4 ................. plug in y = mx+c
x^2 + m^2x^2+2mcx+c^2 = 4
(1+m^2)x^2 + 2mcx + c^2-4 = 0

We have the quadratic function
f(x) = (1+m^2)x^2 + 2mcx + (c^2-4)
where,
x^2 coefficient = (1+m^2)
x coefficient = 2mc
constant = (c^2-4)
If the discriminant is 0, then the quadratic has exactly one root.
This generates exactly one point of intersection between the circle and the tangent line.

d = discriminant
d = (x coefficient)^2 - 4*(x^2 coefficient)*(constant)
d = (2mc)^2 - 4*(1+m^2)(c^2-4)
(2mc)^2 - 4*(1+m^2)(c^2-4) = 0
4m^2c^2 - 4*(1+m^2)(c^2-4) = 0
4( m^2c^2 - (1+m^2)(c^2-4) ) = 0
m^2c^2 - (1+m^2)(c^2-4) = 0/4
m^2c^2 - (c^2-4) - m^2(c^2-4) = 0
m^2c^2 - (c^2-4) - m^2c^2+4m^2 = 0
-(c^2-4)+4m^2 = 0
4m^2 = c^2-4

Answer by greenestamps(13200) (Show Source):
You can put this solution on YOUR website!

The graph of x^2+y^2=4 is a circle with center at the origin and radius 2.

Let O be the origin; let A and B be the y- and x-intercepts of the line y=mx+c; and let C be the point of tangency with the circle.

AOB is a right triangle, and OC is the altitude to the hypotenuse.

The y- and x-intercepts of the given line are

A(0,c)
B(c/m,0)

That gives us the lengths of the legs of triangle AOB as

OA = c
OB = c/m

In right triangle ACO, OA = c and OC = 2, so AC = sqrt(c^2-4).

Triangles AOB and ACO are similar. Set up and solve a proportion using the lengths of the legs of those two triangles.

QED....

SOLUTION: The line y= mx + c is tangent to the curve x^2 + y^2 = 4. Prove that 4m^2 = c^2 - 4