Hard Pre-Calculus problem?

Since the line is tangent to the circle, they have a point in common; call it (x,y).

So, since the y values are the same, we can substitute y = mx + b into the circle equation and simplify:

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

=> expand out and collect like terms

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

This is a quadratic equation, whose solutions are given by the quadratic formula. But, as the hint suggests, since the line is tangent to the circle, there can be only one point in common, so that means there can only be one solution to the above equation for x (otherwise we would have the line intersecting the circle in 2 points).

Thus, a quadratic equation has one solution (with multiplicity 2) if and only if the discriminant (b^2 - 4ac) is zero. Thus, we must have:

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

=> expand things out and divide out by 4

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

=> add b^2 to both sides

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

=> factor out (1 + m^2)

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

=>

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

So the above condition must be true for the line y = mx + b to be tangent to the circle.

x^2+y^2 = r^2 y=mx+b

substituting the line in the circle equation gives

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

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

x^2+m^2 x^2+2mxb+b^2=r^2

grouping terms

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

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

========= ==== =====

A part B part C part

standard equation

is Ax^2 + Bx + C =0

quadratic equation is

-B +/- sqrt( B^2 -4A*C) / 2A

if we apply the quadratic to the above we get

-2mb +/- sqrt( (2mb)^2 -4(1+m^2)*(b^2-r^2) ) /2 * (1+m^2)

-2mb +/- sqrt ( 4m^2 b^2 - 4(b^2-r^2+m^2 b^2 -m^2 r^2)/2 * (1+m^2)

-2mb + 2mb -2b - r +mb - mr / 2 * (1+m^2)

or

-2mb -2mb +2b + r -mb + mr / 2 * (1+m^2)

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

x^2 +m^2x^2 +2mbx +b^2 = r^2

Since line is tangent b^2-4ac=0, so

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

From here it is easilybshown that r^2(1+m^2) = b^2

the derivative is -x/sqrt(r^2-x^2).