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 \(CH\).
+129930 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
