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September 22, 2014

September 22, 2014

Posted by **Allen** on Thursday, October 22, 2009 at 11:17pm.

r^2(1 + m^2) = b^2

Hint:

The quadratic equation x^2 + (mx + b)^2 = r^2 has exactly one solution.

- Alg2 -
**Reiny**, Thursday, October 22, 2009 at 11:35pmsub y = mx + b into x^2 + y^2 = r^2

expanding gave me

x^2 + m^2x^2 + 2bmx + b^2 - r^2 = 0

a^2(1+m^2) + 2bmx + b^2 - r^2 = 0

comparing this to Ax + By + C = 0

A = 1+m^2

B = 2bm

C = b^2 - r^2

we want this quadratic to have only one solution, since there is only one point of contact.

(That was the given hint)

So the discriminant of the quadratic formula must be zero, B^2 - 4AC = 0

(2bm)^2 - 4(1+m^2)(b^2 - r^2) = 0

4b^2m^2 - 4(1+m^2)(b^2 - r^2) = 0

b^2m^2 - (1+m^2)(b^2 - r^2) = 0

b^2m^2 - b^2 + r^2 - b^2m^2 + r^2m^2 = 0

- b^2 + r^2 + r^2m^2 = 0

r^2(1 + m^2) = b^2

as requested.

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