Question

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

Answer 1

$It\ will\ a\ circle:\\\\a\ center:(-2;\ 3)\\\\a\ radius:r=5\\\\the\ equation\ of\ a\ circle:(x-a)^2+(y-b)^2=r^2\\where\ (a;\ b)\ are\ the\ coordinates\ of\ a\ center\ of\ the\ circle.\\\\(x+2)^2+(y-3)^2=5^2\\\\(x+2)^2+(y-3)^2=25\leftarrow answer$

Answer 2

Locus of (x,y) which is at distance 5 from  C (-2,3) will be a circle with C as center and radius 5.
distance between P(x,y)  and C(-2,3) is
        SQUAREROOT [   (x+2)² + (y-3)²  ]  = 5
Now square on both sides (x+2)² + (y-3)² = 25            this is the answer
OR, we can expand the above :  x² +4x + y² -6y -12 = 0