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Question

The number of points P on circle (x3)(x4)+(y1)(y2)=0 such that area of PAB is 1/2 sq units where A (3,2) and B(4,1) is:

A
four
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B
sixteen
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C
two
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D
none of these
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Solution

The correct option is A four
Let P be (h,k)
12∣ ∣hk1321411∣ ∣=12
|h(21)k(34)+1(38)|=1
Taking + sign ( mod will give both + and )
h+k5=1
h+k=6(1)
P is a point on circle.
(h3)(h4)+(k1)(k2)=0
h27h+12+k23k+2=0
h2+k27h3k+14=0(2)
Using (1) and putting k=6h
h2+(6h)27h3(6h)+14=0
h2+36+h212h7h18+3h+14=0
2h216h+32=0
h28h+16=0
(h4)2=0
h=4
k=6h=64=2
(4,2) is the only point on circle which satisfies the given condition on + sign.
On taking sign:
h+k5=1
h+k=4
Putting in (2)
h2+(4h)2+147h3(4h)=0
h2+16+h28h+147h12+3h=0
2h212h+18=0
h26h+9=0
(h3)2=0
h=3
k=4h=1
(3,1) is the point on circle which satisfies the given condition on taking sign.
Two points.
C) Answer.




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