A circle is placed in a 90 degree corner, and that circle is touching both walls of that corner. a point on the circumference of the circle is 18 inches from one wall and 25 inches from the other wall. what is the radius of the circle? if some work could be shown it would be very nice to know for the future.

GatBatNat Apr 15, 2021

#1**+1 **

See the following image :

Let the corner be defined by FGB

Let A be the point on the circumference

Let AC = 18

Let FD =18 - r

Let AD = DE = FB = r

Let AF = 25 - r

Triangle AFD is right such that

FD^2 + AF^2 = AD^2

(18 - r)^2 + ( 25 - r)^2 = r^2 simplify

r^2 - 36r + 324 + r^2 - 50r + 625 = r^2

r^2 - 86r + 949 = 0 we can factor this as

(r - 73) (r - 13) = 0

Setting both factors to 0 and solving for r we have

r = 73 (reject.......r cannot be >18)

r =13 inches

CPhill Apr 15, 2021

#2**+1 **

A slightly different approach....see diagram below:

the angle between the inscribed angle legs of 18 and 25 is 45 degrees

find c using cosine law

c^2 = 18^2 + 25^2 - 2 (18)(25)(cos 45) = 312.604

Now c is also the hypotenuse of the isosceles triangle formed by the two radii

use pythagorean theorem

r^2 + r^2 = c^2

2 r^2 = 312.604

** r = 12.50 inches Hmmmmmm.....not the same as CPhill's answer....but close ! ~ EP**

** OK...I see what I did differently....I went to the points of contact at the wall.....not directly to the walls !!!**

** SEE CPhill's answer ......it is best !!!!!!!!!**

ElectricPavlov Apr 15, 2021