Question: I was wondering if someone could help solve this question,

Show that the point with coordinate$ ( 2 + 2cos (θ) , 2 sin (θ))$ lie on the circle
$x^2 + y^2 = 4x$ and obtain the equation of the tangent to the circle at this point.

The tangent at the points A and B on this circle touch the circle $x^2+y^2=1$ at the points C and D. Find the coordinate of the points of intersection of these tangents, and obtain the equation of the circle through the points A B C D.

the answers are:

$x cos θ + y sin θ = 2 + 2 cos θ $   

(-2,0)  

$x^2+y^2=2x+2$

I can get the equation for the tangent quite easily but can't get the rest, I run into some really messy algebra, by inserting x,in the equation of the tangent, into the formula for the circle, to find the points of intersection. The algebra is horrid and leads to no viable answer.

Thanks for any help.

Answer: 
As pointed out in the comments, your proof is not correct because $(g-f)(g+f)\ge (g(1)+f(1) )(g-f)$ does not necessarily hold for all $x \in [0, \alpha)$. Also I cannot see where the condition $\int\_0^x f \le \int\_0^x g$ for all $x \in [0, 1]$ is used.

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For a proof we can use the [“Second mean value theorem for definite integrals”](https://en.wikipedia.org/wiki/Mean_value_theorem#Mean_value_theorems_for_definite_integrals) in the following form:

> 
> Let $h: [a, b] \to \Bbb R$ be a nonnegative, decreasing function, and $\phi: [a, b] \to \Bbb R$ be an integrable function. Then
> $$
>  \int\_a^b h(x) \phi(x) \, dx = h(a+) \int\_a^c \phi(x) \, dx
> $$
> for some $c \in [a, b]$.
> 
> 
> 

Here $h(a+)$ denotes the one-sided limit $\lim\_{x \to a+} h(x)$.

This can be applied (twice) to our problem: There are $c, d \in [0, 1]$ such that
$$
\begin{align}
\int\_0^1 \bigl(g(x)^2-f(x)^2\bigr)\, dx &= \int\_0^1 g(x)\bigl(g(x)-f(x)\bigr) \, dx
+ \int\_0^1 f(x)\bigl((g(x)-f(x)\bigr) \, dx \\
 &= g(0+) \int\_0^c \bigl(g(x)-f(x)\bigr) \, dx + f(0+) \int\_0^d \bigl(g(x)-f(x)\bigr) \, dx \\
&\ge 0 \, .
\end{align}
$$

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*Remark:* A proof of the second mean value theorem for Lebesgue-integrable functions can be found e.g. in Hobson, E. W. (1909). On the Second Mean-Value Theorem of the Integral Calculus. <https://doi.org/10.1112/plms/s2-7.1.14> .
Let $F(x) = \int\_0^x f(t) \, dt$, and $G(x) = \int\_0^x g(t) \, dt$. Integrating by parts we obtain
$$ \int\_0^1 f(t)^2 \, dt = F(1)f(1) - \int\_0^1 F(t) \, df(t) .$$
Since $df$ is a negative measure, it follows that this is less than or equal to
$$ G(1)f(1) - \int\_0^1 G(t) \, df(t) = \int\_0^1 f(t) g(t) \, dt .$$
Therefore by Cauchy-Schwarz, we have
$$ \int\_0^1 f^2 \le \int\_0^1 fg \le \left(\int\_0^1 f^2\right)^{1/2} \left(\int\_0^1 g^2\right)^{1/2} .$$