+948 A circle is inscribed in a quarter-circle, as shown below. If the radius of the quarter-circle is $4,$ then find the radius of the circle.
+129895 
In right triangle FGB
r^2 +(2 + r)^2 = (4-r)^2
r^2 + r^2 + 4r + 4 = r^2 - 8r + 16
r^2 + 12r - 12 = 0 complete the square on r
r^2 + 12r + 36 = 12 + 36
(r + 6)^2 = 48 take the positive root
r + 6 = sqrt (48)
r = sqrt (48) - 6 = 4sqrt (3) - 6 ≈ .928
+129895 If we extended BF through to the edge of the quarter-circle.....it's length would = 4 ....but we are subtracting off the radius of the circle, r. So BF = 4 - r and it is the hypotenuse of right triangle FGB
Hope that makes sense !!!
+84 so we set a vertex of a triangle at the center of the small circle and another vertex at the center of the square circumcribing the quarter circle and another vertex the corner of the quarter circle. then the triangle then becomes a right triangle since 45+45 is 90 (draw a picture it becomes apparent im just kinda bad at explaning things) so the legs of this right triangle are r*sqrt2 and 2*sqrt2 and the hypotenuse is the hypotenuse of another right triangle with legs of r and r+2. so after squareing all the things according to the pythagorean theoream we get 2r^2+8=2r^2+4r+4. so the radius is \(\boxed{1}\). at least according to my calculationnnssss *nerd emoji*. that took me a while. 😮💨YIKES AFTER SEEING CPHILLS ANSWER I AM VERY UNCONFIDENT OF MY SOLUTION PLEASE TELL ME WHERE I GOT WRONG.
