Let Cᵢ equiv x^2 + y^2 = i^2 (i = 1,2,3) are three circles. If there are 4i points on c

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10-1.Circle and System of Circles

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Let $C_i \equiv x^2 + y^2 = i^2 (i = 1,2,3)$ are three circles. If there are $4i$ points on circumference of circle $C_i$. If no three of all the points on three circles are collinear then number of triangles which can be formed using these points whose circumcentre does not lie on origin, is-

A

$384$

B

$2024$

C

$1360$

D

$1744$

Solution

There are $4,8$ and $12$ points on circles ${C_1},{C_2}$ and ${C_3}$ respectively for required triangle not all three point should be concyclic.

$\therefore $ required number of triangle $ = {\,^{24}}{C_3} – {\,^4}{C_3} – {\,^8}{C_3} – {\,^{12}}{C_3} = 1744$

Standard 11

Mathematics

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