As shown in the figure, circle $C_1$ with center $O_1$ and radius $r_1$ touches the square $VWXY$ at points $P$ and $Q$ while circle $C_2$ with center $O_2$ and radius $r_2$ touches the square $VWXY$ at points $R$ and $S$. The two circles touch each other at $T$.
Given $r_1 = 1$ cm and $\overline{VY} = \overline{VW} = 4$ cm, $r_2 =$ cm.

Question

A rectangle $ABCD$ has sides $AB = 45 \, \text{cm}$ and $BC = 26 \, \text{cm}$. Point $E$ is the midpoint of side $CD$. Find the radius of the incircle of the triangle $\triangle AED$.

Answer 1

Look at the horizontal and vertical gaps between the two circle centers directly, without setting up a full coordinate grid.

Distance of $O_1$ from the two sides at $W$: circle $C_1$ sits in the corner at $W$, touching the two sides meeting there, so its center is $r_1 = 1$ cm away from each of those two sides.
Distance of $O_2$ from the two sides at $Y$: circle $C_2$ sits in the corner at $Y$, touching the two sides meeting there (VY and XY), so its center is $r_2$ cm away from each of those sides.
Horizontal and vertical separation between centers: since the square has side 4 cm, the horizontal distance between $O_1$ and $O_2$ is $4 - r_2 - r_1 = 4 - r_2 - 1 = 3 - r_2$. Because both circles sit tucked into opposite corners along the same diagonal of the square, the vertical separation works out to the same value, $3 - r_2$.
Applying Pythagoras: the straight-line distance between the two centers is the hypotenuse of a right triangle with both legs equal to $3 - r_2$:
$$O_1O_2 = \sqrt{(3-r_2)^2 + (3-r_2)^2} = (3-r_2)\sqrt{2}$$
Using external tangency: since the circles just touch each other at $T$ from outside, the center distance must equal $r_1 + r_2 = 1 + r_2$. Setting the two expressions equal:
$$(3-r_2)\sqrt{2} = 1+r_2$$
Solving for $r_2$: expand and collect terms:
$$3\sqrt2 - r_2\sqrt2 = 1+r_2$$ $$3\sqrt2 - 1 = r_2(1+\sqrt2)$$ $$r_2 = \frac{3\sqrt2-1}{1+\sqrt2}$$ Multiply top and bottom by $(\sqrt2-1)$ to clear the surd from the denominator:
$$r_2 = (3\sqrt2-1)(\sqrt2-1) = 6-3\sqrt2-\sqrt2+1 = 7-4\sqrt2$$

Numerically, $7-4\sqrt2 \approx 7 - 5.657 = 1.343$ cm, a sensible radius that is bigger than $r_1=1$ cm, matching the figure where $C_2$ is clearly the bigger circle, and smaller than the square itself.

Let's summarize:

Both circle centers sit at a fixed distance, their own radius, from the two square sides they touch.
The gap between centers along both directions works out to $3 - r_2$, so Pythagoras gives the straight-line center distance.
Setting that equal to $r_1+r_2$ (external tangency) and solving gives $r_2 = 7 - 4\sqrt2$ cm.

So $r_2 = 7 - 4\sqrt2$ cm, which is option (C).

Answer 2

Look at the horizontal and vertical gaps between the two circle centers directly, without setting up a full coordinate grid.

Distance of $O_1$ from the two sides at $W$: circle $C_1$ sits in the corner at $W$, touching the two sides meeting there, so its center is $r_1 = 1$ cm away from each of those two sides.
Distance of $O_2$ from the two sides at $Y$: circle $C_2$ sits in the corner at $Y$, touching the two sides meeting there (VY and XY), so its center is $r_2$ cm away from each of those sides.
Horizontal and vertical separation between centers: since the square has side 4 cm, the horizontal distance between $O_1$ and $O_2$ is $4 - r_2 - r_1 = 4 - r_2 - 1 = 3 - r_2$. Because both circles sit tucked into opposite corners along the same diagonal of the square, the vertical separation works out to the same value, $3 - r_2$.
Applying Pythagoras: the straight-line distance between the two centers is the hypotenuse of a right triangle with both legs equal to $3 - r_2$:
$$O_1O_2 = \sqrt{(3-r_2)^2 + (3-r_2)^2} = (3-r_2)\sqrt{2}$$
Using external tangency: since the circles just touch each other at $T$ from outside, the center distance must equal $r_1 + r_2 = 1 + r_2$. Setting the two expressions equal:
$$(3-r_2)\sqrt{2} = 1+r_2$$
Solving for $r_2$: expand and collect terms:
$$3\sqrt2 - r_2\sqrt2 = 1+r_2$$ $$3\sqrt2 - 1 = r_2(1+\sqrt2)$$ $$r_2 = \frac{3\sqrt2-1}{1+\sqrt2}$$ Multiply top and bottom by $(\sqrt2-1)$ to clear the surd from the denominator:
$$r_2 = (3\sqrt2-1)(\sqrt2-1) = 6-3\sqrt2-\sqrt2+1 = 7-4\sqrt2$$

Numerically, $7-4\sqrt2 \approx 7 - 5.657 = 1.343$ cm, a sensible radius that is bigger than $r_1=1$ cm, matching the figure where $C_2$ is clearly the bigger circle, and smaller than the square itself.

Let's summarize:

Both circle centers sit at a fixed distance, their own radius, from the two square sides they touch.
The gap between centers along both directions works out to $3 - r_2$, so Pythagoras gives the straight-line center distance.
Setting that equal to $r_1+r_2$ (external tangency) and solving gives $r_2 = 7 - 4\sqrt2$ cm.

So $r_2 = 7 - 4\sqrt2$ cm, which is option (C).

Show Hint

The Correct Option is C

Solution and Explanation

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