The locus of a point which is at a distance of 4 units from ( 3 , - 2 ) in x y -plane is

The locus of a point which is at a distance of 4 units from (3,-2) in xy-plane is
  1. x2+y2+6x-4y+16=0
  2. x2+y2-6x-4y+3=0
  3. x2+y2-6x+4y-16=0
  4. x2+y2-6x+4y-3=0

Solution

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

Asked in: AP EAMCET 2021 (20 Aug Shift 1)

Practice more Hyperbola questions on Aicharya