In circle O as shown above, AB=2 cm and DC=6 cm, ∠BAC=∠ACD=45∘. What is the radius of circle O?
+23246 Let V be the midpoint of arc(AB) and Z the midpoint of arc(CD).
Draw VWXYZ, where W is the point of intersection with AB (W will be the midpoint of AB);
X is the point of intersection with AC;
Y is the point of intersection with CD (Y will be the midpoint of CD).
VWXYZ will be perpendicular to both AB and CD.
Since AB = 2, AW = 1 and WX = 1.
Since CD = 6, CY = 3 and XY = 3.
Call XW = a and call YZ = b.
Since WXYZ is a diameter and AB is a chord of the circle: AW · WB = VW · WZ.
---> 1 · 1 = a · (1 + 3 + b) ---> 1 = a · (4 + b) ---> 4a + ab = 1 ---> ab = 1 - 4a
Since WXYZ is a diameter and CD is a chord of the circle: CY · YD = YZ · VY.
---> 3 · 3 = b · (1 + 3 + a) ---> 9 = b · (4 + a) ---> 4b + ab = 9 ---> ab = 9 - 4b
Combining these two results: 1 - 4a = 9 - 4b ---> 4b - 4a = 8 ---> b - a = 2
---> b = 2 + a
Since a · (4 + b) = 1 ---> a · (4 + (2 + a) ) = 1 ---> a · (6 + a) = 1
---> 6a + a2 = 1 ---> a2 + 6a - 1 = 0
---> a = -3 + sqrt(10) [quadratic formula]
Since b · (4 + a) = 9 ---> b · (4 + (b - 2) ) = 9 ---> b · (2 + b) = 9
---> 2b + b2 = 9 ---> b2 + 2b - 9 = 0
---> b = -1 + sqrt(10) [quadratic formula]
The diameter will be: (a) + (1) + (3) + (b) = ( -3 + sqrt(10) ) + (1) + (3) + ( -1 + sqrt(10) )
= 2sqrt(10)
So, the radius is: sqrt(10)
