Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
The locus of a point which is at a distance of 4 units from (3,-2) in xy-plane is
MathematicsStraight LinesAP EAMCETAP EAMCET 2021 (20 Aug Shift 1)
Options:
  • A x2+y2+6x-4y+16=0
  • B x2+y2-6x-4y+3=0
  • C x2+y2-6x+4y-16=0
  • D x2+y2-6x+4y-3=0
Solution:
2614 Upvotes Verified Answer
The correct answer is: x2+y2-6x+4y-3=0

Let the point be h,k

Distance between the two points will be =h-32+k+22

4=h-32+k+22

Squaring both sides we get,

16=h2+9-6h+k2+4k+4

h2-6h+k2+4k-3=0

Therefore, required locus is x2+y2-6x+4y-3=0

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.