Rectangle at the corner .

Using Pythagoras' Theorem

for the right-angled triangle in the figure

$(r - a)^2 + (r - b)^2 = r^2$

Now, a = 1 cm and b = 2 cm. So,

$(r - 1)^2 + (r - 2)^2 = r^2$ or

$r^2 - 2r + 1 + r^2 - 4r + 4 = r^2$ .
This simplifies to

$r^2 - 6r + 5 = 0.$ On factorizing, we get

$(r - 5)(r - 1) = 0$ .

Thus, the radius of the circle is 5 cm.

(neglecting 1 as a possible solution)

We can also solve this as an analytic geometry problem. The small rectangle can be placed in the bottom-left corner of the square, so we can think at the circumference as if it was a circumference placed in the first quadrant and tangent to the axis and we can interpret the rectangle as formed by a point P(2;1) on the circumference and his projections on the axis. A general circumference equation is $x^2 + y^2 + ax + by + c = 0$ . Since this circumference is tangent to the axis, his center has the same value for both the abscissa and the ordinate; then $a = b$ . The equation becomes $x^2 + y^2 + ax + ay + c = 0$ . We can force the passage for $P$ a nd we can find $c$ in function of $a$ : $4 + 1 + 2a + a + c = 0 c = -5 - 3a$ .

The equation becomes

$x^2 + y^2 + ax + ay - 5 - 3a = 0$ .

Since it is tangent to the y-axis, we can substitute 0 with x in the equation and determinate the value of a that makes the discriminant 0. We get

$y^2 + ay - 5 - 3a = 0$ .

The determinant is

$a^2 + 4(5 + 3a)$

And it is 0 when $a = -2$ or when $a = -10$ .

The radius in a circumference is $r = sqrt{(-a/2)^2 + (-b/2)^2 - c}$ . Substituting b and c with $a$ and $-5 - 3a$ we get $sqrt{2 * (-a/2)^2 + 5 + 3a}$ .

For $a = -2$ we get $r = 1$ For $a = -10$ we get $r = 5$

For $r = 1$ the point P is not at the bottom left of the circumference, then

$radius = 5 cm$ .