The number of points belonging to the set $A = {(2, - 4), (5, - 4), (4, - 5), (1, - 3), (2, - 11), (2, - 5)}$ lying inside the circle $x^{2} + y^{2} - 4 x + 12 y + 15 = 0$ and below the line $2 x - 3 y = 16,$ is

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Solution

$L = - 2 x + 3 y + 16 = 0, S = x^{2} + y^{2} - 4 x + 12 y + 15 = 0 at (5, - 4), (4, - 5), (2, - 5)$

$L_{1} < 0 and S_{1} < 0$

$\Rightarrow$ Number of such points is three

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