The curves $x^{2} + y^{2} + 6 x - 24 y + 72 = 0$ and $x^{2} - y^{2} + 6 x + 16 y - 46 = 0$ intersect at 4 points. If the sum of the distances of the four points from the point (-3,2) is 10m, then m=?

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answer is 4.

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Detailed Solution

(4) ${(x + 3)}^{2} + {(y - 12)}^{2} = 81$
${(x + 3)}^{2} - {(y - 8)}^{2} = - 9$
Let the point of intersection be $p (α, β)$
$\begin{array}{l} {(α + 3)}^{2} + {(β - 12)}^{2} = 81 - (1) \\ {(α + 3)}^{2} - {(β - 8)}^{2} = - 9 - (2) \end{array}$
Add (1) and (2), then $α^{2} + 6 α - 4 β + 13 = 0$
subtract (2) from (1) $β^{2} - 20 β + 59 = 0$
$\Rightarrow d = \sqrt{{(α + 3)}^{2} + {(β - 2)}^{2}} \Rightarrow β$
sum of the distance of all intersection points $= 2 (y_{1} + y_{2}) = 2 (β_{1} + β_{2}) = 2 \times 20 = 40$