The radii of two concentric circles are 16 cm and 10 cm. #AB# is a diameter of the bigger circle. #BD# is tangent to the smaller circle touching it at #D#. What is the length of #AD#?

Adopting the origin #(0,0)# as the common center for #C_i# and #C_e# and calling #r_i=10# and #r_e=16# the tangency point #p_0=(x_0,y_0)# is at the intersection #C_i nn C_0# where

#C_i->x^2+y^2=r_i^2#
#C_e->x^2+y^2=r_e^2#
#C_0->(x-r_e)^2+y^2= r_0^2#

here #r_0^2 = r_e^2-r_i^2#

Solving for #C_i nn C_0# we have

#{(x^2+y^2=r_i^2),((x-r_e)^2+y^2=r_e^2-r_i^2):}#

Subtracting the first from the second equation

#-2xr_e+r_e^2=r_e^2-r_i^2-r_i^2# so

#x_0 = r_i^2/r_e# and #y_0^2= r_i^2-x_0^2#

Finally the sought distance is

#bar(AD)=sqrt((r_e+x_0)^2+y_0^2)=sqrt(r_e^2+3r_i^2)#

or

#bar(AD)=23.5797#

Explanation:

If #bar(BD)# is tangent to #C_i# then #hat(ODB) = pi/2# so we can apply pythagoras:

#bar(OD)^2+bar(DB)^2=bar(OB)^2# determining #r_0#

#r_0^2= bar(OB)^2-bar(OD)^2=r_e^2-r_i^2#

The point #D# coordinates, called #(x_0,y_0)# should be obtained before calculating the sought distance #bar(AD)#

There are many ways to do that. An alternative method is

#y_0=bar(BD)sin(hat(OBD))# but #sin(hat(OBD))=bar(OD)/bar(OB)#

then

#y_0 = sqrt(r_e^2-r_i^2)(r_i/r_e)# and
#x_0=sqrt(r_i^2-y_0^2)#

As per given data the above figure is drawn.

O is the common center of two concentric circles

#AB->"diameter of the bigger circle"#

#AO=OB->"radius of the bigger circle"=16 cm#

#DO->"radius of the smaller circle"=10cm#

#BD->"tangent to the smaller circle"->/_BDO=90^@#

Let #/_DOB=theta=>/_AOD=(180-theta)#

In #Delta BDO->cos/_BOD=costheta=(OD)/(OB)=10/16#

Applying cosine law in #Delta ADO# we get

#AD^2=AO^2+DO^2-2AO*DOcos/_AOD#

#=>AD^2=AO^2+DO^2-2AO*DOcos(180-theta)#

#=>AD^2=AO^2+DO^2+2AO*DOcostheta#

#=>AD^2=AO^2+DO^2+2AO*DOxx(OD)/(OB)#

#=>AD^2=16^2+10^2+2xx16xx10xx10/16#

#=>AD^2=556#

#=>AD=sqrt556=23.58cm#