1. Find the centre and radius of $C_1$.

Given: $x^2+y^2-2(x+y)+1=0$

i.e. $x^2+y^2-2x-2y+1=0$

Complete squares: $x^2-2x+y^2-2y+1=0$ $(x-1)^2-1+(y-1)^2-1+1=0$ $(x-1)^2+(y-1)^2=1$

So, the centre of $C_1$ is $O_1=(1,1)$ and radius is $1$.

2. Write the equation of $C_2$.

$C_2$ has centre $(-1,0)$ and radius $2$, so $(x+1)^2+y^2=4$ Expanding, $x^2+y^2+2x-3=0$

3. Find the common chord (radical axis) of $C_1$ and $C_2$.

The common chord is obtained by subtracting the equations of the circles.

From $C_1$: $x^2+y^2-2x-2y+1=0$ From $C_2$: $x^2+y^2+2x-3=0$

Subtracting second from first: $(-2x-2y+1)-(2x-3)=0$ $-4x-2y+4=0$ $2x+y-2=0$

Hence the line of common chord is $2x+y-2=0$

4. Find the point where this line intersects the $y$-axis.

On the $y$-axis, $x=0$. So, $2(0)+y-2=0 \Rightarrow y=2$

Thus, $P=(0,2)$

5. Find the square of the distance of $P$ from the centre of $C_1$.

Centre of $C_1$ is $(1,1)$. So, $OP^2=(0-1)^2+(2-1)^2=1+1=2$

Therefore, the required square of the distance is $\boxed{2}$

6. Check with options.

• A: $4$ ❌
• B: $6$ ❌
• C: $2$ ✅
• D: $1$ ❌

So the correct option is C.