Question:

To find the equation of the common chord of circles $C_1$ and $C_2$, we first need to understand that the common chord is the line segment where the two circles intersect. The equation of circle $C_1$ is given by $x^2+y^2-2(x+y)+1=0$, and $C_2$ has its centre at $(-1,0)$ with a radius of 2. The equation of $C_2$ can be written as $(x+1)^2 + y^2 = 4$. Step 1: The centre of $C_1$ can be found by converting its equation into the standard form of a circle, $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the centre. For $C_1: x^2+y^2-2(x+y)+1=0$, completing the square gives $$ \begin{aligned} x^2+y^2-2(x+y)+1 &= 0 \\ x^2 - 2x + 1 + y^2 - 2y + 1 &= 1 \\ (x-1)^2 + (y-1)^2 &= 1 \end{aligned} $$ Thus, the centre of $C_1$ is $(1,1)$. The centre of $C_2$ is given as $(-1,0)$. Step 2: The slope of the line joining these two centres is $\frac{1-0}{1-(-1)} = \frac{1}{2}$. The slope of the common chord, being perpendicular, is the negative reciprocal of $\frac{1}{2}$, which is $-2$. Step 3: However, instead of directly using the slope to find the equation of the common chord, we can use the given equation of the common chord, which is $2x+y=2$. This equation represents a line, and we are told that this line is the common chord of $C_1$ and $C_2$. To find where this line intersects the $y$-axis, we set $x=0$ in the equation $2x+y=2$. Step 4: Substituting $x=0$ into $2x+y=2$ gives $y=2$. Therefore, the point $P$ where the common chord intersects the $y$-axis is $(0,2)$. Step 5: To find the square of the distance of $P$ from the centre of $C_1$, we use the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$, where $(x_1, y_1)$ is the centre of $C_1$, which is $(1,1)$, and $(x_2, y_2)$ is the point $P$, which is $(0,2)$. Step 6: Substituting these values into the formula gives $$ \begin{aligned} d^2 &= (0 - 1)^2 + (2 - 1)^2 \\ &= (-1)^2 + (1)^2 \\ &= 1 + 1 \\ &= 2 \end{aligned} $$ Step 7: Given the provided options and reevaluating the calculation with the correct approach: the centre of $C_1$ is $(1,1)$ and point $P$ is $(0,2)$, the correct step to find $d^2$ is indeed $d^2 = (1-0)^2 + (1-2)^2 = 1 + 1 = 2$. This matches option 1. Therefore: $\boxed{1}$ <div class="key-concept"><strong>Key Concept:</strong> Common chord: $S_1-S_2=0$. $S_1=x^2+y^2-2x-2y+1$, $S_2=x^2+y^2+2x-3=0$. Common chord: $-4x-2y+4=0\Rightarrow2x+y=2$.</div> <div class="trap-box"><strong>Trap:</strong> At $y$-axis ($x=0$): $y=2$, so $P=(0,2)$. Centre of $C_1=(1,1)$. $d^2=(1-0)^2+(1-2)^2=2$.</div>