Question:

Step 1: Identify the given equations of the curves. The two given curves are: Equation 1: $C_1: x^2 + y^2 + 6x - 24y + 72 = 0$ Equation 2: $C_2: x^2 - y^2 + 6x + 16y - 46 = 0$ Step 2: Find the equations that define the intersection points by combining the given equations. To find the coordinates of the intersection points, we add and subtract the two equations. Adding Equation 1 and Equation 2: $$ (x^2 + y^2 + 6x - 24y + 72) + (x^2 - y^2 + 6x + 16y - 46) = 0 $$ $$ 2x^2 + 12x - 8y + 26 = 0 $$ Divide by 2: $$ x^2 + 6x - 4y + 13 = 0 $$ Complete the square for the $x$ terms: $$ (x^2 + 6x + 9) - 9 - 4y + 13 = 0 $$ $$ (x+3)^2 - 4y + 4 = 0 $$ $$ (x+3)^2 = 4(y-1) \quad (*) $$ Subtracting Equation 2 from Equation 1: $$ (x^2 + y^2 + 6x - 24y + 72) - (x^2 - y^2 + 6x + 16y - 46) = 0 $$ $$ 2y^2 - 40y + 118 = 0 $$ Divide by 2: $$ y^2 - 20y + 59 = 0 \quad (**) $$ Step 3: Solve the quadratic equation for $y$ to find the $y$-coordinates of the intersection points. The quadratic equation for $y$ is $y^2 - 20y + 59 = 0$. Let the roots of this equation be $y_1$ and $y_2$. By Vieta's formulas, the sum of the roots is $y_1 + y_2 = -(-20)/1 = 20$. The product of the roots is $y_1 y_2 = 59/1 = 59$. Since both the sum and product are positive, both roots $y_1$ and $y_2$ must be positive. For each $y$ value ($y_1$ and $y_2$), from equation $(*)$, we have $(x+3)^2 = 4(y-1)$. Since $y_1, y_2 > 0$ and $y_1, y_2 \ne 1$ (as $1^2 - 20(1) + 59 = 40 \ne 0$), we have $y-1 > 0$. Thus, for each $y_i$, there are two distinct values for $x$: $x+3 = \pm 2\sqrt{y_i-1}$. This means there are four distinct intersection points. Let these points be $(x_j, y_j)$ where $j=1,2,3,4$. Two points will have $y$-coordinate $y_1$, and the other two will have $y$-coordinate $y_2$. Step 4: Express the distance from an intersection point $(x,y)$ to the point $P(-3,2)$ in terms of $y$. Let $(x,y)$ be any of the four intersection points, and let $P_0 = (-3,2)$. The distance $d$ between $(x,y)$ and $P_0$ is given by the distance formula: $$ d = \sqrt{(x - (-3))^2 + (y - 2)^2} = \sqrt{(x+3)^2 + (y-2)^2} $$ From equation $(*)$ in Step 2, we know that $(x+3)^2 = 4(y-1)$. Substitute this into the distance formula: $$ d = \sqrt{4(y-1) + (y-2)^2} $$ $$ d = \sqrt{4y - 4 + y^2 - 4y + 4} $$ $$ d = \sqrt{y^2} $$ Since we established in Step 3 that the $y$-coordinates ($y_1, y_2$) are positive, we have $\sqrt{y^2} = |y| = y$. So, the distance from any intersection point $(x,y)$ to $P_0(-3,2)$ is $d=y$. Step 5: Calculate the sum of these distances for all four intersection points. Let the four intersection points be $(x_{1a}, y_1)$, $(x_{1b}, y_1)$, $(x_{2a}, y_2)$, and $(x_{2b}, y_2)$. The sum of the distances from these four points to $P_0(-3,2)$ is: $$ \sum_{j=1}^4 d_j = y_1 + y_1 + y_2 + y_2 $$ $$ \sum_{j=1}^4 d_j = 2y_1 + 2y_2 = 2(y_1 + y_2) $$ From Step 3, we know that $y_1 + y_2 = 20$. Therefore, the sum of the distances is $2(20) = 40$. Step 6: Equate the sum of distances to $10k$ and solve for $k$. The problem states that the sum of distances from these four points to the point $(-3,2)$ is $10k$. We found the sum of distances to be $40$. So, we have: $$ 10k = 40 $$ $$ k = \frac{40}{10} $$ $$ k = 4 $$ The final answer is $\boxed{4}$.