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Question

From the point P(16,7), tangents PQ and PR are drawn to circle x2+y2-2x-4y-20=0.

If C is the centre of the circle, then the area of equilateral PQCR is


A

450 sq units

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B

15 sq units

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C

50 sq units

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D

75 sq units

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Solution

The correct option is D

75 sq units


Explanation for the correct option:

Step 1. Find the area of PQCR:

Given: the equation of the circle is

x2+y2-2x-4y-20=0

(x-1)2+(y-2)2=20+12+22

(x+g)2+(y+f)2=g2+f2-c

So, radius of the circle r=g2+f2-c

=1+4+20

=25

=5

Given the coordinates of external point P=(16,7)

and the coordinates of the center C =(1,2)

Step 2. Find the value of PC:

PC2=(16-1)2+(7-2)2 [Distance formula =(x2-x1)2+(y2-y1)2]

=152+52

=225+25

=250

PR2=PQ2 [Tangents from an external point to the circle are equal]

=PC2-r2 [using Pythagoras theorem]

=250-25

=225

PR=PQ

=15

Step 3. Area of PCQ or PCR =12×b×h

=12×15×5

=752 sq. units

Area of PQCR =2×Area of PCQ or PCR

=2×752

=75 sq. units.

Hence, Option ‘D' is Correct.


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