The graph of $x^{2} + y^{2} + 6 x - 24 y + 72 = 0$ and $x^{2} - y^{2} + 6 x + 16 y - 46 = 0$ intersect at four points. Sum of the distance of these four points from point $(- 3, 2)$ be $α,$ then $\frac{α}{20}$ =_____________.

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Detailed Solution

$x^{2} + y^{2} + 6 x - 24 y + 72 = 0 \dots \dots . (i)$

$x^{2} - y^{2} + 6 x + 16 y - 46 = 0 \dots \dots \dots (i i)$

(i)+ (ii) , $2 x^{2} + 12 x - 8 y + 26 = 0, x^{2} + 6 x - 4 y + 13 = 0$ Question Image

${(x + 3)}^{2} = 4 (y - 1) \dots \dots \dots (i i i)$

Pt of intersecting also satisfy equation………(iii)

$\sum_{i = 1}^{4} \sqrt{{(x i + 3)}^{2} + {(y i - 2)}^{2}} = \sum_{i = 1}^{4} \sqrt{4 (y i - 1) + {(y i - 2)}^{2}} \sum | y i |$

Similarly equation also satisfy $2 y^{2} - 40 y + 118 = 0, y^{2} - 20 y + 59 = 0 \Rightarrow \sum_{i = 1}^{4} y i = 2 \times 20 = 40$

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