Circle in which side of square is given

Question

I am not Able to proceed with this question .

Answer 1

how can PQ be independent of a? I am getting .6 a

Answer 2

I get that the length of "?" is 3a/5.

This is probably the longest possible way, so here goes.

Let the two points on the left and right ends of the line labled "?" be P and Q.

Put the origin at the base of the circle near "a".

Let $\angle DAP = \angle ADQ =g$, and let $\sin g = s, \cos g = c, \tan g = t $.

The equation of $AP$ is $y = t(a/2-x) $ and the equation of $BQ$ is $y = t(x+a/2) $.

These intersect when $x = 0$ at $(0, at/2)$. This is the center of the circle. Call this $R$.

The distance $AP = a$, and

$\begin{array}\\ AR^2 &=(a/2)^2+(ta/2)^2\\ &=(a/2)^2(1+t^2)\\ &=(a/2)^2(1+(s/c)^2)\\ &=(a/2)^2(1/c)^2\\ \text{so}\\ AR &=a/(2c) \end{array} $

Therefore the radius of the circle is $a-a/(2c) =a(1-1/(2c)) $.

But this has to be $at/2$, so $at/2 =a(1-1/(2c)) $ or $t =2-1/c $ or $s/c = 2-1/c$ or $s = 2c-1$.

Squaring, $4c^2-4c+1 =s^2 =1-c^2 $, so $5c^2 = 4c$ so $c = 0$ or $c = 4/5$.

From this, $s = \sqrt{1-16/25} =3/5$, and, indeed, $s = 2c-1$.

Also, $t = s/c = 3/4$.

If point $P$ is at $(-u, v)$, then $(u+a/2)^2+v^2 = a^2 $ and $v/(u+a/2) =t$ or $v = t(u+a/2) $ so

$\begin{array}\\ a^2 &=(u+a/2)^2+t(u+a/2)^2 \\ &=(u+a/2)^2(1+t^2)\\ &=(u+a/2)^2(1/c^2)\\ \text{so}\\ ac &=u+a/2\\ \text{or}\\ u &=a(c-1/2)\\ &=3a/10\\ \end{array} $

Finally, the length of $?$ is $2u = 6a/10 =3a/5 $.