Question

The equation of locus of the point which is at a
distance 5 units from A(4,-3) is

Answer 1

Answer:

Let an arbitrary point be P(x,y)

Then the distance of the point P from the point A(−2,3) is

d=

(x+2)

2

+(y−3)

2

d

2

=(x+2)

2

+(y−3)

2

Now the distance is given to be 5,

Hence

(x+2)

2

+(y−3)

2

=25

x

2

+y

2

+4x−6y+4+9=25

x

2

+y

2

+4x−6y+13=25

x

2

+y

2

+4x−6y−12=0.

Answer 2

Answer:

x^2+y^2-8x+6y=0

Step-by-step explanation:

√(x-4)^2+(y+3)^2=5

(x-4)^2+(y+3)^2=25

x^2+16-8x+y^2+9+6y=25

x^2+y^2-8x+6y=0