Chords AB and CD of a circle are parallel, and angle BDC = angle ACD = 45 degrees. If CD = 10 and the radius of the circle is 6, then find AB.
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Let M be the midpoint of the chord of 10
Let the center of the circle = (0,0) = O
The equation of the circle is x^2 + y^2 = 36 (1)
Connect OC, OM
OC =6, MC =5
We can find OM thusly
sqrt ( OC^2 -MC^2) = sqrt (6^2 -5^2) = sqrt (11)
Let the coordinates of C = (5, -sqrt (11) )
The slope of the line containing segment AC = -1
The equation of this line is y = -(x -5) -sqrt (11)
y+ x = 5 -sqrt (11) square both sides
x^2 + 2xy + y^2 = 36 - 10sqrt (11) sub (1) into this
36 + 2xy = 36 -10 sqrt (11)
2xy = -10sqrt (11)
xy = -5sqrt (11)
y = -5sqrt (11)/x
Sub this into (1) to find the x cooordinate of A
x^2 + ( -5sqrt (11)/x)^2 =36
x^2 + 275/x^2 = 36 multiply through by x^2
x^4 + 275 = 36x^2
x^4 - 36x^2 + 275 = 0
Factor as
(x^2 - 25)(x^2 -11) = 0
The solutions are x = -5, 5, -sqrt (11), sqrt (11)
The x coordinate of A cannot be -5 ...so the x coordinate of A = -sqrt (11)
And due to symmetry, the x coordinate of B =sqrt (11)
So
AB = sqrt (11) - -sqrt (11) = 2sqrt (11)
