Find the radius of the circle inscbribed inside the square. The distance from the side of the square to circle is (2,1)

Question

Find the radius of the circle inscbribed inside the square. The distance from the side of the square to circle is (2,1)see the attached image

Answer 1

Assuming the center at the origin we have $$x^2+ y^2=r^2$$ with the point $(-r+2, -r+1)$ on the circle.

Plugging in and simplifying the equation we get $$r^2-6r+5=0$$

The solution $r=5$ is acceptable.

Answer 2

Asuming the square's sides are parallel to the $\;x\,-$ axis and the $\;y\,-$ axis, and since the left lower vertex is at $\;(0,0)\;$ and the circle's equation is

$$(x-r)^2+(y-r)^2=r^2\;\;\;\text{(why?)}$$

we then get from the fact that $\;(2,1)\;$ is on the circle that

$$(2-r)^2+(1-r)^2=r^2\implies r=\ldots$$

Finish the argument

Answer 3

Using the Pythagoras theorem:
$a=1$ $cm$
$b=2$ $cm$
$(r-a)^2+(r-b)^2=r^2$
$(r-1)^2+(r-2)^2=r^2$
$r^2+1-2r+r^2+4-4r=r^2$
$r^2-6r+5=0$
$r=5$ $cm$
$r=1$ $cm$
The $r=5$ $cm$ is the acceptable answer.