Perpendicular cords AC and DE intersect at B so that the lengths of AB,BC and BD are 3, 4 and 2 respectively. Find the radius of the circle.
+129899 By the intersecting chords theorem
EB * DB = AB * CB
EB * 2 = 3 * 4
EB = 6
Angle EBC =(1/2) (minor arc EC + minor arc AD)
90 = (1/2) ( minor arc EC + minor arc AD)
180 = minor arc EC + minor arc AD
EC^2 = 4^2 + 6^2 = 52
AD^2 = 3^2 + 2^2 = 13
Using the Law of Cosines
EC^2 = 2 r^2 - 2r^2 cos A
AD^2 = 2r^2 - 2r^2 cos (180 - A) where A is a central angle....simplify
Note.....cos (180 - A) = -cos A
So....the second equation becomes
AD^2 = 2r^2 - 2r^2 (-cos A)
13 = 2r^2 + 2r^2 cos A
So we have
52 =2r^2 -2r^2 cos A
13 = 2r^2 +2r^2 cos A add these equations
65 = 4r^2
65/4 = r^2
r = sqrt (65) / 2
Here's a different method.
Take a line parallel to DE, a half a unit to the right.
It's the perependicular bisector of AC so will pass through the centre of the circle.
Drop a perpendicular from the centre of the circle onto DE, bisecting it.
Then, by Pythagoras, r^2 = 4^2 + (1/2)^2 = 16 + (1/4) = 65/4, so r = sqrt(65)/2.
