Question

Find the equation to the locus of the point which is at a distance of $5$ unit from $(-2,3)$ in a plane.

Answer 1

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

$PQ=\sqrt{{(h+2)}^2+{(k-3)}^2}{\left(PQ\right)}^2={(h+2)}^2+{(k-3)}^2=>25={(h+2)}^2+{(k-3)}^2$

For locus replace $h\stackrel{}{\to }x,k\stackrel{}{\to }y=>{(x+2)}^2+{(y-3)}^2=25$ is a circle with radius $5$ and centre $(-2,3)$