Smallest circle possible

Connecting the edges of the outer most points, we get a triangle with sides $2$ , $\sqrt5$ and $\sqrt5$ (by Pythagoras's Theorem). Now the problem is to find the circumradius of this triangle using the following formula. $R=\frac{abc}{\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}}$ $R=\frac{2 \cdot \sqrt5 \cdot \sqrt5}{\sqrt{(2+\sqrt5+\sqrt5)(2+\sqrt5-\sqrt5)(2+\sqrt5-\sqrt5)(\sqrt5+\sqrt5-2)}}$ $R=\frac{2 \cdot5 }{\sqrt{(2\sqrt5+2)(2)(2)(2\sqrt5-2)}}$ $R=\frac{10}{\sqrt{(20 - 4)(2)(2)}}$ $R=\frac{10}{\sqrt{64}}$ $R=1.25$

Let the endpoints of the T be the points $(1,0), (-1,0), (0,-2)$ . We are looking for a circle with equation $x^2+(y-y_0)^2=r^2$ passing through these points. Filling in the coordinates of the known points, we get two equations:

$1+y_0^2=r^2$ and $0+(-2-y_0)^2=r^2$ , from which follow:

$1+y_0^2=4+4y_0+y_0^2$

$0=3+4y_0$

$y_0=-\frac{3}{4}$

Using this value for $y_0$ for any point we find $r^2=\frac{25}{16}$ , so that

$r=\frac{5}{4}=\boxed{1.25}$ .

Call the endpoints of the top line segment A and B, and the point at the bottom of the figure C. We are looking for the radius of the circumcircle of $\triangle ABC$ . Let D be the midpoint of $\overline{AB}$ , E the midpoint of $\overline{BC}$ and D the midpoint of $\overline{AC}$ . Construct the perpendicular bisectors of $\overline{BC}$ and $\overline{AC}$ , intersecting at G on $\overline{CD}$ . G is the center of the circumcircle of $\triangle ABC$ . See the diagram below.

Notice that