Question

The equation of the common tangent touching the circle (x-3)^2 + y^2 = 9 and the parabola y^2 = 4x above the x axis.

Answer 1

Let us find the tangent to the parabola \( y = 2 \sqrt x\) at a point \( (m, 2\sqrt m)\) then determine m so this tangent is also tangent to the circle

Answer 2

It is easy to find it using the slope and a point, you get
\[
g(x)=\frac{m+x}{\sqrt{m}}
\]

Answer 3

Now substitute the point (x,g(x)) in the circle to find the point of intersections of the tangent and the circle, you get
\[
\left(\frac{m+x}{\sqrt{m}}\right)^2+(x-3)^2-9=\\\frac{x
^2}{m}+m+x^2-4 x=\\\left(\frac{1}{m}+1\right)
x^2-4 x + m=0
\]

Answer 4

Now
\[
\left(\frac{1}{m}+1\right)
x^2-4 x + m=0
\]
is a quadratic in x. To be tangent, we need the two roots to be equal, so we need
\[b^2-4 a c=0=16-4 \left(\frac{1}{m}+1\right) m=12-4
m=0
\]
Hence m=3

Answer 5

Finally the equation of the tangent is
\[
g(x)=\frac{m+x}{\sqrt{m}}=\frac{3+x}{\sqrt{3}}
\]

Answer 6

In the attached graph the red line is the common tangent and the blue graph is that of the
parabola. @satellite73 @experimentX @ganeshie8

Answer 7

May be the people, I tagged above can do it in an easier way.

Answer 8

Your answer is correct. But what's wrong with my method?When I equate and square the second terms of both. Why doesn't that give the correct ans?