Let the circle $$C_1 : x^2 + y^2 - 2(x + y) + 1 = 0$$ and $$C_2$$ be a circle having centre at $$(-1, 0)$$ and radius 2. If the line of the common chord of $$C_1$$ and $$C_2$$ intersects the $$y$$-axis at the point $$P$$, then the square of the distance of $$P$$ from the centre of $$C_1$$ is :

Given $$C_1: x^2 + y^2 - 2(x+y) + 1 = 0$$ and $$C_2$$ has centre $$(-1, 0)$$ and radius 2.

We rewrite $$C_1$$ as $$ x^2 - 2x + y^2 - 2y + 1 = 0 $$ which becomes $$ (x-1)^2 - 1 + (y-1)^2 - 1 + 1 = 0 $$ and hence $$ (x-1)^2 + (y-1)^2 = 1. $$

The centre of $$C_1$$ is $$(1, 1)$$ and the radius is 1.

For $$C_2$$ with centre $$(-1,0)$$ and radius 2 the equation is $$ (x+1)^2 + y^2 = 4, $$ which expands to $$ x^2 + 2x + 1 + y^2 = 4 $$ and simplifies to $$ x^2 + y^2 + 2x - 3 = 0. $$

The common chord is obtained by subtracting the equations of $$C_2$$ from $$C_1$$: $$ (x^2 + y^2 - 2x - 2y + 1) - (x^2 + y^2 + 2x - 3) = 0, $$ giving $$ -2x - 2y + 1 - 2x + 3 = 0, $$ or $$ -4x - 2y + 4 = 0 $$ which simplifies to $$ 2x + y - 2 = 0. $$

Setting $$x=0$$ in the chord equation yields $$2(0)+y-2=0$$, so $$y=2$$, and hence $$P=(0,2)$$.

The square of the distance from $$P$$ to the centre of $$C_1$$ is $$ d^2 = (0-1)^2 + (2-1)^2 = 1 + 1 = 2. $$

The correct answer is Option (1): 2.