Inside And Outside

Relevant wiki: Circles Problem Solving - Basic

Let the radius of bigger circle and smaller circle is $a$ and $b$ .

So, shaded area= $\pi a^2-\pi b^2=\dfrac{25\pi}{2}\Rightarrow \color{#3D99F6}{a^2-b^2}=\dfrac{25}{2}$

Let the contact point of tangent to smaller circle be $F$ .

In $\triangle FOQ$ .

$OQ^2=OF^2+FQ^2$

$a^2=b^2+FQ^2$

$\color{#3D99F6}{a^2-b^2}=FQ^2$

$\dfrac{25}{2}=FQ^2$

$FQ=\dfrac{5\sqrt{2}}{2}$

And we know that $2FQ=PQ$ .

$\therefore PQ=2×\dfrac{5\sqrt{2}}{2}=\boxed{5\sqrt{2}}$

Area of annulus = $\pi r_l^2 - \pi r_s^2 = \pi(r_l^2 - r_s^2) = \frac{25\pi}{2}$ . So $r_l^2 - r_s^2 = \frac{25}{2}$ .

We draw a right triangle with vertices at the centre of the circle(s), the tangent point, and Q. Let's call the angle at the centre of the circle $\theta$ .

Immediately we can see that the length of the adjacent side to that angle is the radius of the smaller circle, $r_s$ . So we can say that $r_l\cos\theta = r_s$ (because the hypotenuse of our right triangle is the radius of the large circle, $r_l$ ). We substitute this expression for $r_s$ back into our first equation:

$r_l^2 - (r_l\cos\theta)^2 = r_l^2(1-\cos^2\theta) = \frac{25}{2} \space \space \space \Longrightarrow \space \space \space r_l^2\sin^2\theta = \frac{25}{2}$ (trigonometric identity).

So $r_l\sin\theta = \frac{5}{\sqrt{2}}$ , and we recognize this as being the opposite side to $\theta$ ! We also notice that this value represents only half of the length of PQ. So the answer is $2\times\frac{5}{\sqrt{2}} = 5\sqrt{2}$ .

From the figure, the area of the annulus is equal to $\left(\dfrac{PQ}{2}\right)^2 \pi$ , so

$\left(\dfrac{PQ}{2}\right)^2 \pi=\dfrac{25 \pi}{2}$

$PQ = \sqrt{50} = \color{#69047E}\large{\boxed{5 \sqrt(2)}}$ $\color{#3D99F6}\boxed{\text{answer}}$

$A_{yellow}=\pi(R^2-r^2)$

$\dfrac{25}{2}\pi = \pi (R^2-r^2)$

$\dfrac{25}{2}=R^2-r^2$

By pythagorean theorem,

$\left(\dfrac{PQ}{2}\right)^2=R^2-r^2$

$\dfrac{PQ^2}{4}=\dfrac{25}{2}$

$PQ^2=50$

$PQ=\sqrt{50}=\sqrt{25(2)}=\sqrt{5^2(2)}=$ $\boxed{5\sqrt{2}}$