Squares in circle

In the figure, $AE^{2}+DE^{2}=r^{2}$ $\Rightarrow$ $(x+4)^{2}+4=r^{2}$ and $(12-x)^{2}+36=r^{2}$ ............ $(x+4)^{2}+4=(12-x)^{2}+36$ $\Rightarrow$ $x=5$ . $\Rightarrow$ $r^{2}=\sqrt{9^{2}+4}$ .
$\therefore$ $r=\boxed{\sqrt{85}}$ .

By law of sines,

4root17 / sin theta = 2R sin theta = 2 / root5

so, 4 root17 = 4R / root 5,

R = root 85

Radius $R$ of circle inscribing two squares with sides $a$ & $b$ & sharing a common edge by the following general formula

$\boxed{R=\frac{\sqrt{5(5a^2+5b^2+6ab)}}{8}}$

As per given question, setting $a=\sqrt{16}=4$ & $b=\sqrt{144}=12$ in above general formula we get

$R=\frac{\sqrt{5(5(4)^2+5(12)^2+6\cdot 4\cdot 12)}}{8}=\frac{\sqrt{5\times 1088}}{8}=\sqrt{\frac{5\times 1088}{64}}=\sqrt{85}$

Let the circle have radius $r$ and be centered at the origin. Without loss of generality, orient the chord so that it is horizontal and lies along the line $y = d$ for a real number $d \gt 0.$

The equation of the circle is $x^{2} + y^{2} = r^{2}.$ The smaller square will be inscribed in the portion of the circle above the line $y = d.$ As this square will have side lengths of $4$ units and will be symmetric about the $y$ -axis, the upper right corner of the square will have coordinates $(2, 4 + d).$ As this point lies on the circle, we have that

$4 + (4 + d)^{2} = r^{2},$ (i).

The larger square lies below $y = d$ and has side lengths of $12$ units, and so the lower right corner of this square has coordinates $(6, d - 12).$ As this point lies on the circle, we have that

$36 + (d - 12)^{2} = r^{2},$ (ii).

Comparing (i) and (ii), we see that

$4 + (4 + d)^{2} = 36 + (d - 12)^{2} \Longrightarrow 20 + 8d + d^{2} = 180 + d^{2} - 24d$

$\Longrightarrow 32d = 160 \Longrightarrow d = 5.$

Plugging this value for $d$ into (i) gives us that $4 + 9^{2} = r^{2} \Longrightarrow \boxed{r = \sqrt{85}}.$

We are essentially looking for the circumradius of a triangle with area $96$ and side lengths $12,4\sqrt{17},8\sqrt{5}$ . By Euler's formula $R=\frac{abc}{4S}$ the outcome is $\sqrt{85}$ .

Let the given chord be $MN$ . Let the center of the circle be $O$ and its radius be ' $r$ '.

Let the bigger square be $ABCD$ and the smaller square be $PQRS$ such that points $A, B$ and $R, S$ lies on $MN$ .

Draw circle's diameter joining the centers of $ABCD$ and $PQRS$ . Say it intersect the chord $CD$ at $C^{'}$ and chord $PQ$ at $P^{'}$ .

$|C^{'}P^{'}| = |DA| + |SP| = 12 + 4 = 16$

Also,

$|C^{'}P^{'}| = |C^{'}O| + |OP^{'}| = \sqrt{r^2 -(\frac{12}{2})^2} + \sqrt{r^2 - (\frac{4}{2})^2}$

$\Rightarrow \sqrt{r^2 -(\frac{12}{2})^2} + \sqrt{r^2 - (\frac{4}{2})^2} = 16$

$\Rightarrow \sqrt{r^2 - 2^2} = 16 - \sqrt{r^2 -6^2}$

$\Rightarrow r^2 - 4 = 256 + r^2 -36 - 32 \sqrt{r^2 -6^2}$

$\Rightarrow 32 \sqrt{r^2 -6^2} = 256 + 4 -36$

$\Rightarrow \sqrt{r^2 -6^2} = 7$

$\Rightarrow r^2 -6^2 = 7^2$

$\Rightarrow r = \sqrt{85}$