A circle is moving in which at one instance the circle touches the point \((-2,3)\) and tangent to the line \(x=6\).

To calculate the locus of the center of the circle, equate the distance between the center of the circle to the distance (*perpendicular*) between the center of the circle and the line.

Let \((x,y)\) be the center of the circle. The distance between the center of the circle to the point \((-2,3)\) is \(d^2 =(x+2)^2+ (y-3)^2\) and the distance between the center of the circle to the line is \(d = \frac{1(x) + 0(y) - 6 }{\sqrt{1^2+0}} = x - 6\)

\((x+2)^2 + (y-3)^2 = (x-6)^2\)

\(x^2 +4x+4+ y^2 -6y +9 = x^2 -12x +36\)

\( y^2 -6y + 16x-23 =0\) Answer

Distance between two points \((x,y) ~ and ~(x,_1,y_1)\): \(d^2 = (x_1-x)^2+(y_1-y)^2 \)

Distance between a point \((x_o, y_o)\) and a line \((Ax + By + C = 0)\): \(d = \frac{Ax_o + By_o +C}{\sqrt{A^2+B^2}}\)