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In a circle with center \(O\) and radius \(5\), \(AB\) is a diameter.  Points \(C\) and \(H\) on the circumference are such that \(CB=1.25\) and \(OH\) and \(CB\) are parallel .  Find the length of \(CH\). Jan 14, 2021

#2
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Line segment CH = 7.5 units (!!!) Jan 14, 2021

#1
0   Jan 14, 2021
edited by CPhill  Jan 14, 2021
#3
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CB = 1.25          ∠CBO ≅ ∠HOB

∠COB = 2(sin-1(BM / OB))

∠CBO = ∠HOB = 1/2(180 - ∠COB)

∠COH = ∠COB + ∠HOB

CH = 2*[sin(COH /2) * OH] jugoslav  Jan 14, 2021
edited by jugoslav  Jan 15, 2021
#2
+1

Line segment CH = 7.5 units (!!!) jugoslav Jan 14, 2021
#4
+1

A little different from jugoslav's method  : C = (5.0)

Construct circles  x^2 + y^2  =25      and  (x -5)^2 + y^2 = 1.25^2

Find the coordinates of B

Subtract the  first equation  from the  second

(x-5)^2  - x^2  = 1.25^2 - 25

x^2  -10x + 25  - x^2  =   1.25^2 -25

-10x =   1.25^2 - 50

-10x =  25/16  - 800/16

-10x =  -775/16

x= 775/160  = 155/32

And

(155/32)^2  + y^2 = 25

y ^2 =  25 - (155/32)^2

y ^2  =  1575/1024       take the negative  root

y  =  sqrt  (1575/1024)=  - (15/32)sqrt (7)

So B  =( 155/32,  -(15/32) sqrt (7)  )

Slope of line segment BC =   ( 0  + 15sqrt 7 (32) )  /( 5 - 155/32)  =  [ (15sqrt (7)) / 32 ] /  (5/32)   =

3sqrt (7) =   sqrt ( 63)

The equation of  a line through (0,0)  with this slope has the equation

y= sqrt (63)x

The intersection of this line with   x^2 + y^2  = 25    has the x coordinate

x^2 +  63x^2   =  25

64x^2  = 25

x^2  = 25/64       take the negative root

x = - 5/8

And

y = -sqrt  (25 - 25/64)  = - sqrt (1575/64)  = -( 15/8) sqrt (7)

H = ( -5/8, -(15/8) sqrt (7) )

So  CH  = sqrt  [ (5  + 5/8)^2  + ( (15/8)sqrt (7) )^2 ] =

sqrt  [( 45/8)^2  +  1575/64  ]  =

sqrt  [ 45^2 + 1575 ] / 8

sqrt  [ 3600] / 8  =

60/8   =

7.5   Jan 15, 2021