Help finding a centre of a circle

Why not just solve the quadratic? Collect like terms in $x$ as follows. Move everything to one side to get $$(x-a)^2 + 4x^2 - 16 = 0.$$ Then expand the square: $$begin{align} 0 &= x^2 - 2ax + a^2 + 4x^2 - 16 \ &= 5x^2 - 2ax + a^2 - 16. end{align}$$ Then using the quadratic formula gives you $$x = frac{2a pm sqrt{(-2a)^2 - 4(5)(a^2-16)}}{2(5)} = frac{a pm 2sqrt{20-a^2}}{5}.$$ This means the intersection of the line and the circle will have two solutions if $a^2 - 20 > 0$, no solutions if $a^2 - 20 < 0$, and exactly one solution if $a^2 - 20 = 0$. Since the line is supposed to be tangent, $a$ must satisfy the last case, hence $a = 2sqrt{5}$. We take the positive value for $a$ since we know the circle's center is on the positive $x$-axis.

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Alternatively, we can exploit the geometry of similar right triangles. Let $P = (x_0, 2x_0)$ be the point of tangency of the line $y = 2x$ with the circle, and let the origin be $O = (0,0)$ and the center of the circle $A = (a,0)$. Let $Q = (x_0, 0)$ be the foot of the perpendicular from $P$ to the $x$-axis. Then we have $$triangle PQO sim triangle APO,$$ hence $$frac{AP}{PO} = frac{PQ}{QO} = frac{2x_0}{x_0} = 2.$$ Since $AP = 4$, the radius of the circle, it follows that $PO = 2$, and by the Pythagorean theorem, $$a = AO = sqrt{AP^2 + PO^2} = sqrt{4^2 + 2^2} = sqrt{20} = 2 sqrt{5}.$$

Since the center of the circle lies on the $x$-axis, the circle touches the lines $y=2x, y=-2x$, which are equally inclined from the $x$-axis, at the same $x$ and therefore, $text{Discriminant}left((x-a)^2+4x^2=16right)=0Rightarrow4a^2=4cdot5(a^2-16)Rightarrow a=2sqrt5$.

The approach given by Tavish is the easier method. But this Alternate method is going along the steps you have already taken.

since $y=2x$ is a tangent to the circle $(x-a)^2+y^2=16$ on solving both the equations simultaneously we should get repeated roots.

i.e., the quadratic(in $x$) $(x-a)^2+(2x)^2=16$ has $D=0$ (Discriminant of the quadratic)

Guide:

The angle between $y=2x$ and the $x$-axis is $theta$ where $tan theta = 2$.

Let the intersection between $y=2x$ and the radius be $P$. Let the center be $A$. Consider $triangle OAP$,

We have $$sin theta = frac{4}{a}$$

I will leave the task of finding $sin theta$ to you.

Hint:

The distance of the center of the circle from the lines $y=pm2x$ is $4$. Use the fact that the distance between the point $(x_0,y_0)$ and the line $ax+by+c=0$ is $$left | frac{ax_0+by_0+c}{sqrt{a^2+b^2}} right|$$