Two mutually perpendicular chords AB and CD meet at a point P inside the circle such that AP=6 units PB=4 units and DP=3 units. What is the area of the circle?

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Solution

The correct option is A

As AB and CD are two chords that intersect at P, AP $$ PB = CP $$ PD
$6 * 4 = C P * 3 \Rightarrow C P = 8$
From centre O draw OM $⊥$ AB and ON $⊥$ CD
We get AM = MB = 5 (perpendicular from center bisects the chord)
Therefore MP = 1, ON = 1, CD = 11
CN = ND = 5.5 (perpendicular from centre bisects the chord)
$O N^{2} + C N^{2} + O C^{2}$
$1 + (5.5)^{2} = r^{2}$
Area of circle = $π r^{2} = π (1 + (5.5)^{2}) = \frac{125 π}{4}$

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