Question

Find the locus of the point whose shortest distance from the circle $x^2+y^2-2x+6y-6=0$is equal to its distance from the line $x-3=0$

Answer 1

Hint : First note that shortest distance from the circle is the distance from its centre minus the radius. Hence the required locus is

$\sqrt{\left[\right(x-1{)}^2+(y+3{)}^2]}-4+x-3$

or $(x-1{)}^2+(y+3{)}^2=(x-1{)}^2$

or $y^2+6y-4x+9=0$