Find the radius of circle, given tangents from three vertices of a square

Solution 1:

HINT...let the coordinates of the centre of the circle be $(x,y)$ with the bottom left-hand corner of the square being the origin. Then, considering distances from the origin and from other corners of the square to the centre of the circle, you have: $$x^2+y^2=64+r^2$$ $$x^2+(10-y)^2=9+r^2$$ and $$(10-x)^2+(10-y)^2=49+r^2$$

You can solve these simultaneously to get $x$ and $y$ and hence get $r$.

Solution 2:

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 : $$r^2+8^2=3^2+(10-r)^2$$