Two mutually perpendicular chords AB and CD intersect at P. AP = 4, PB = 6, CP = 3. Find the radius of the circle.

- 31.25
^{(1/2)} - 37.5
^{(1/2)} - 26
^{(1/2)} - 52
^{(1/2)}

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 = ( OX^{2} + XB^{2} )^{1/2} OB = 31.25^{1/2} Thus radius = 31.25^{1/2} The radius can also be found out using the triangle YOD.

The question is **"Find radius of the circle."**