Question

find the equation of a point which is at a distance of 5 from (3,2)​

Answer 1

Let the point be P(h,k) , the fixed point be Q(−2,3) and PQ=5

PQ=

(h+2)

2

+(k−3)

2

(PQ)

2

=(h+2)

2

+(k−3)

2

=>25=(h+2)

2

+(k−3)

2

For locus replace h

x,k

y=>(x+2)

2

+(y−3)

2

=25 is a circle with radius 5 and centre (−2,3)