a) r 2 (1+m 2 ) = b 2 [The quadratic equation x 2 + (mx + b) 2 = r 2 has exactly

Question

a) r2 (1+m2) = b2 [The quadratic equation x2 + (mx + b)2 = r2 has exactly one solution.

b) The point of tangency is (-r2m/b , r2/b)

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

The tangent line to a circle may be defined as the line that intersects in a single point, called the point of tangency. See the figure. 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

Explanation / Answer

I'll not give you the very complete answer because it's necessary for you to think on your own a little bit:)



The equation of the circle x^2+y^2=r^2 shows that the center of the circle is at (0,0).
a)According to the definition of the tangent line, the line only intersects the circle in a single point, i.e. these simultaneous equations x2+y2 = r2 and y=mx+b has exactly 1 solution. (I don't know how to enter a bracket, sorry).

Let y in x2+y2 = r2 be mx+b. Then we know that x2 + (mx + b)2 = r2 has only 1 solution. (m2+1)x2+2mbx+b2-r2=0 has only one solution.

As it has 1 solution and m2+1>0, we know that the of this equation equals to 0.

=2mb*2mb-4*(m2+1)(b2-r2)=0, then you can figure it out easily:).

b)We just need to figure out the root of the equation above.

x=(-2mb±0)/2(m2+1).

As m2+1=b2/r2(shown in the part a)

x=-2mb/2(b2/r2)=-mr2/b

Then y can be easily figured out using y=mx+b. (You still need to use r2 (1+m2) = b2 to get y in this form)



c) You should already know that if the two lines are perpendicular, the product of their slopes equals to -1.

Suppose the tangent intersects the circle at point M.

Then the slope of OM is (r*r/b-0)/(-r*r*m/b-0)=-1/m

Then the product of these two slopes is (-1/m)*m=-1

As the product of these two slopes is -1, these two lines are perpendicular.

This is the end of the proof:)

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