Posted by Dealie on .
Hi, I am studying for a precalculus quiz and I do not understand this hw problem:
"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,
a. 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^2r^2
4m^2b^2 = 4(1+m^2)(b^2r^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,
Thank you so much. you totally rock