Let C_i equiv x^2 + y^2 = i^2 for i = 1, 2, 3 | vedclass

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(D) The number of points on circles $C_1, C_2,$ and $C_3$ are $4(1)=4, 4(2)=8,$ and $4(3)=12$ respectively. Total number of points $n = 4 + 8 + 12 = 2

Vedclass · Accepted Answer

$1744$

Vedclass · Answer

(D) The number of points on circles $C_1, C_2,$ and $C_3$ are $4(1)=4, 4(2)=8,$ and $4(3)=12$ respectively. Total number of points $n = 4 + 8 + 12 = 24$. The total number of triangles that can be formed using $24$ points is $^{24}C_3$. $A$ triangle will have its circumcentre at the origin if and only if all three vertices lie on the same circle (i.e.,they are concyclic). Number of triangles with circumcentre at the origin = (Triangles formed by points on $C_1$) + (Triangles formed by points on $C_2$) + (Triangles formed by points on $C_3$) = $^4C_3 + ^8C_3 + ^{12}C_3$. Required number of triangles = Total triangles - Triangles with circumcentre at the origin. Required number = $^{24}C_3 - (^4C_3 + ^8C_3 + ^{12}C_3) = 2024 - (4 + 56 + 220) = 2024 - 280 = 1744$.

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