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Question

The locus of the point which moves so that the square of its distance from the point (3,−2) is numerically equal to its distance from the line 5x−12y=13 can be

A
x2+y2+14+83x64y13=0
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B
x2+y2+14=83x64y13
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C
x2+y2+12+73x40y13=0
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D
x2+y2+12=73x40y13
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Solution

The correct option is D x2+y2+12=73x40y13
Let the variable point be (h,k)
According to the given condition, we get
(h3)2+(k+2)2=|5h12k13|13

Now
Case 1: Point (h,k) and origin lies on the same side of the line 5x12y13=0
(h3)2+(k+2)2=(5h12k13)1313(h2+k2)+16978h+52k=5h+12k+1313(h2+k2+12)73h+40k=0
Therefore, the locus is
x2+y2+12=73x40y13

Case 2: Point (h,k) and origin lies on opposite sides of the line 5x12y13=0
(h3)2+(k+2)2=(5h12k13)1313(h2+k2)+16978h+52k=5h12k1313(h2+k2+14)83h+64k=0
Therefore, the locus is
x2+y2+14=83x64y13

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