Friday

April 18, 2014

April 18, 2014

Posted by **Dealie** on Saturday, August 28, 2010 at 11:57pm.

"The tangent line to a circle may be defined as the point that intersects a circle in a single point... If the equation of the circle is x^2+y^2=r^2 and the equation of the tangent line is y=mx+b show that:

a. r^2(1+m^2)=b^2 [Hint: the quadratic equation x^2+(mx+b)^2=r^2 has exactly one solution]

b.the point of tangency is (-r^2m/b,r^2/b)

c. the tangent line is perpendicular to the line containing the circle and the point of tangency

- Math (Pre Cal) -
**drwls**, Sunday, August 29, 2010 at 12:26ama. For the equation

x^2+(mx+b)^2 = r^2

to have one solution, the discriminant of the quadratic equation

Ax^2 + Bx + C = 0

must be zero

B^2 = 4 AC

A = 1+m^2

B = 2mb

C = b^2-r^2

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

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

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

b. x = -B/(2A) = -mb/(1+m^2)

= -mb[r^2/b^2] = -mr^2/b

c. Don't you mean the line containing the CENTER OF the circle and the point of tangency?

- Math (Pre Cal) -
**Dealie**, Sunday, August 29, 2010 at 9:46amThank you so much. you totally rock

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