Two mutually perpendicular chords AB and CD intersect at P. AP = 4, PB = 6, CP = 3. Find radius of the circle.
Two mutually perpendicular chords AB and CD intersect at P. AP = 4, PB = 6, CP = 3. Find the radius of the circle.
When 2 chords AB and CD intersect at P then AP * PB = CP * PD
Hence 4 * 6 = 3 * PD
Thus, PD = 8
Now AB = AP + PB = 10
And CD = CP + PD
Thus, CD = 11
Consider the circle with center O.
Drop a perpendicular from O to chord AB and CD.
This will bisect the chords at X and Y i.e AX=XB and CY = YD.
Here AX = AP + PX i.e 5 = 4 + PX
PX = 1
Similarly, since CD = 11, PY+CP+YD = 11,
=> PY = 11-3-5.5 = 2.5.
PY = OX and PX = OY.
So, PXOY will from a rectangle as seen in the figure.
Now consider the triangle BOX, it is a right triangle where OB is the radius.
XB = 5, OX = 2.5
Then OB = ( OX2 + XB2 )1/2 OB = 31.251/2 Thus radius = 31.251/2 The radius can also be found out using the triangle YOD.
The question is "Find radius of the circle."