If the small circle is inscribed in the sector, I assume it is tangent to the larger circle, as well as the radii of the sector. In that case, let
O = center of large circle
P = center of inscribed circle
r = radius of inscribed circle
OP = 5-r
r^2+r^2 = (5-r)^2
r = 5(√2 - 1)
A quarter-circle with radius $5$ is drawn. A circle is drawn inside the sector, which is tangent to the sides of the sector, as shown. Find the radius of the inscribed circle.
The rest is in asymptote
size( 100 ) ;
// Not to scale. Don't try anything here, kids
pair O = origin ;
draw( circle((-1,1),1) ) ;
pair I = intersectionpoint( arc((-1,1),1,90,180) , O--4dir(135) ) ;
real radius = length( I-O ) ;
pair A = radius*dir(180) ;
pair B = radius*dir(90) ;
draw( O--arc(O,radius,90,180)--cycle ) ;
label( "$O$" , O , SE ) ;
label( "$A$" , A , SW ) ;
label( "$B$" , B , NE ) ;
1 answer
1 answer