Arithmetic mean of the length of tangents to a circle

Question

There are eight circles $S (i = 1,2,3,.....8)$ of radius $1,2,3.......8$ respectively and all the circles are passing through $(8,8)$ and $(9,9)$ then the arithmetic mean of the length of tangents drawn from $(0,0)$ to these circles is:
A)$12$
B)$6$
C)$3$
D)$1$

What I have tried:

As the length of $PQ$ is smaller than all the diameters, it must be a chord. Now by parametric equation, the coordinates of $B(x_1,y_1)$ can be found out as: $$\dfrac{x_1-\frac{17}{2}}{-\frac1{\sqrt2}}=\dfrac{y_1-\frac{17}{2}}{\frac1{\sqrt2}}=\pm \sqrt{r^2+\frac1{\sqrt2}}$$

Now using $B$, I found $OB$, then using it, I found $OA$

$$OB=\sqrt{\left(\dfrac{17}{2}\right)^2+\dfrac{r^2}{2}+\dfrac1{2\sqrt2}}\\OA=\sqrt{\left(\dfrac{17}{2}\right)^2+\dfrac{3r^2}{2}+\dfrac1{2\sqrt2}}$$

I am struck. I can't calculate all the individual lengths of tangents and take their mean. There must be another way.

Answer 1

Note that $O(0,0)$, $P(8,8)$ and $Q(9,9)$ are collinear.

By the tangent-secant theorem or the power of point theorem, at point $O$,

$$\begin{align*} OA^2 &= OP \cdot OQ\\ &= 8\sqrt2 \cdot 9\sqrt 2\\ &= 144\\ OA &= 12 \end{align*}$$

So the length of each tangent is $12$, regardless of the radius.

Answer 2

If I continue with your method, it's possible if I make two corrections on how you applied Pythagoras' theorem:

Firstly in $\triangle BDQ$, about the length $BD$, which should instead satisfy:

$$\begin{align*} BD^2 + \left(\frac{1}{\sqrt2}\right)^2 &= r^2\\ BD &= \sqrt{r^2 -\frac{1}{2}}\\ \end{align*}$$

So the length of $OB$ obtained from your method should be:

$$\begin{align*} x_1 &= \frac{17}{2} \mp \sqrt{\frac{r^2}{2} -\frac{1}{2\cdot2}}\\ y_1 &= \frac{17}{2} \pm \sqrt{\frac{r^2}{2} -\frac{1}{2\cdot2}}\\ OB &= \sqrt{x_1^2 + y_1^2}\\ &= \sqrt{2\left[\left(\frac{17}{2}\right)^2 + \frac{r^2}{2} - \frac{1}{2\cdot2}\right]}\\ &= \sqrt{2\left(\frac{17}{2}\right)^2 + r^2 - \frac{1}{2}}\\ \end{align*}$$

Secondly in $\triangle OAB$, $\angle A$ is right-angled, not $\angle B$ as in your attempt. So

$$\begin{align*} OA^2 + r^2 &= OB^2\\ OA &= \sqrt{\left[2\left(\frac{17}{2}\right)^2 + r^2 - \frac{1}{2}\right] - r^2}\\ &=\sqrt{\frac{17^2}{2} - \frac{1}{2}}\\ &=\sqrt{\frac{18\cdot 16}{2}}\\ &= 12 \end{align*}$$

So the length of each tangent is $12$, regardless of the radius.