The above is the image of the question. The brown, blue, green lines are tangent to the circle and the measure of a side of the square is 10. The question is to find the value of the radius.
My Attempt:
I took O as the centre of the circle and drew radius to the tangent point of each tangent. Then I found the lengths of the lines joining the circumcenter and each vertex of the square. But I didn't find any useful relations among the attempts I tried.
So anyone in this community could help me with this question.
Sorry for the bad writing in the picture.
Thank you!
It is an interesting setup and we can try to understand the geometry of it.
Let us first drop the diagram on the coordinate plane. The square has vertices $A(0,0), B(10,0),C(10,10),D(0,10)$. The given circle has center $O(h,k)$.
Take circles at $A,C,D$ of radius $8,7,3$ respectively. The tangent lengths indicate that these three circles cut the given circle at right angles. Hence the center of given circle is meeting point of the common chords/tangents (also known as radical center) of the three new circles.
Notice that $7+3=10$. This means circles $\odot(C),\odot(D)$ are tangent on the side of the square and their common tangent (radical axis) is at $x=3$. So $h=3$.
To compute the $y$-coordinate, we can find the equation of common chord of circles $\odot(A),\odot(D)$. This is found by equating the equations of $\odot(A),\odot(D)$.
$$x^2+y^2-8^2=x^2+(y-10)^2-3^2$$ $$\Rightarrow y=31/4 =k$$
Hence center of purple circle is at $(3,31/4)$. It follows that its radius is $$10-31/4=9/4=\boxed{2.25}$$
Alternatively, as @MathLover says, once we realized that the purple circle is tangent to side $CD$ of the square, we can apply Pythagoras to write $AO^2$ in two ways : $$r^2+8^2=3^2+(10-r)^2$$