The circle $x^2 + y^2 + 6x - 24y + 72 = 0$ and $x^2 - y^2 + 6x + 16y - 46 = 0$ intersect at four points. The sum of distances from these four points to the point $(-3, 2)$ is $10k$. Then value of $k$ is equal to ___.
Step-by-Step Solution
Key Concept: Use algebraic manipulation of circle/curve equations to find intersection points and then apply distance formula with the constraint on sum of ordinates.
Adding and subtracting the two circle equations $(x+3)^2 = 4(y-1)$ and $y^2 - 20y + 59 = 0$ gives the intersection points. From the constraint that $y_1 + y_2 = 20$ and $y_1, y_2 > 0$, we find the sum of distances from each intersection point to the respective curve equals $\sum\sqrt{(y_i+3)^2 + (y_i-2)^2} = \sum\sqrt{4(y_i-1)+(y_i-2)^2} = \sum y_i = 2(y_1+y_2) = 40$.
Correct Answer: 4