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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.

 

 Jun 23, 2024
 #1
avatar+129735 
+1

 

 

 

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

 

cool cool cool

 Jun 23, 2024
edited by CPhill  Jun 23, 2024
 #3
avatar+85 
+1

i tried doing that but i wasnt sure that the 4-r (like the quarter circle radius would connect to the center of the circle)would be a straight line can you explain to me how to be sure

shmewy  Jun 23, 2024
edited by shmewy  Jun 23, 2024
 #4
avatar+129735 
+1

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 !!!

 

cool cool cool

CPhill  Jun 23, 2024
 #5
avatar+85 
+1

right but how are you sure that when you extend it it will form a nice straight line all the way to B? I get everything else :)

shmewy  Jun 23, 2024
 #6
avatar+129735 
+1

See the new image

 

The line containing BF will   intersect the quarter-circle at H

 

So  BH =  the radius of the quarter-circle = 4

 

And FH  is  just the radius of our small  circle = r

 

So

 

BH  -  FH  =  4 - r  =  BF

 

 

cool cool cool

CPhill  Jun 23, 2024
edited by CPhill  Jun 23, 2024
 #2
avatar+85 
+1

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.

 Jun 23, 2024
edited by shmewy  Jun 23, 2024
edited by shmewy  Jun 23, 2024

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