Question

find the equation of locus of a point which is at a distance of 5 units from (-2 , 3 ) xoy plane​

Answer 1

Answer:

the equation is

x²+y²+4x-6y-12= 0

Answer 2

Answer:

x²+y²+4x-6y-12=0

Step-by-step explanation:

The equation for the palne is

(x-a)²+(y-b)²= c²

here

a=-2

b=3

c= distance =5

Given point is (-2,3)

Let the coordinates of the moving point be (x,y)

Distance of (x,y) from (-2,3) is

sqrtof (x-(-2))²+(y-3)²

sqrtof (x+2)²+(y-3)²

as mentioned in the given problem it's at a distance of 5 units.

equating the above equation to the distance

sqrtof (x+2)²+(y-3)²=5

SOBS

(x+2)²+(y-3)² =5²

x²+4x+4+y²-6y+9=25

x²+y²+4x-6y-12=0