Question 1132209
<pre>
Draw in radii OK, OH and OL and let their lengths be r.
Let OA = 4, the perpendicular from O to HK.
Let the red line OB = x, the perpendicular from O to HL, the length
of which is what we are to find.

Triangle OHK is isosceles, so the perpendicular from O to HK, which 
is 4, bisects the base HK.  HK is given as 16, so AH = AK = 8.

Similarly, triangle OHL is isosceles, so the perpendicular from O to HL,
which is our unknown x, bisects the base HL.  HL is given as 10, so 
BH = BL = 5.

{{{drawing(400,400,-10,10,-10,10,

circle(0,0,sqrt(80)),
line(0,0,sqrt(80),0), line(0,0,-5.366563146,7.1555417528),
line(0,0,-8.645710761,-2.29165561), 

line(-8.645710761,-2.29165561,-5.366563146,7.1555417528),
line(-5.366563146,7.1555417528,sqrt(80),0),
red(line(0,0,-7.006166101,2.431859098)),line(0,0,1.788854382,3.57770876),
locate(9.1,.4,K),locate(-5.5,8,H),locate(-9.3,-2.1,L),locate(0,0,O),
locate(1.7,4.4,A),locate(-7.7,3,B), locate(4.6,0,r),locate(1,2,4),
locate(4.5,3,8),locate(-1.2,5.9,8),locate(-6.2,5,5),locate(-7.7,.6,5), locate(-4,2.2,x)locate(-4.8,-1,r),locate(-2.3,4,r)  
)}}} 

We use the Pythagorean theorem on either right triangle OAH or OAK:

r² = 8²+4² = 64+16 = 80
 
We use the Pythagorean theorem on either right triangle OBH or OBL:

r² = x²+5², and we have already found r² = 80, so

80 = x²+25

55 = x²

{{{sqrt(55)=x}}}

So the solution is {{{sqrt(55)}}}

Edwin</pre>