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Lines, triangles, quadrilaterals and circles from 7 points | JEE Advanced

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Question
77 points are given in a plane (as shown in the below figure): points 11, 22, 33 lie on one straight line and points 33, 44, 55, 66, 77 lie on another straight line, the two lines meeting at point 33. Then using these 77 points Question attachment
ANumber of lines that can be formed is 1010.correct
BNumber of triangles that can be formed is 2424.correct
CNumber of quadrilateral that can be formed is 66.correct
DMaximum number of circles that can be formed is 2424, where each circle contains atleast three of 77 pointscorrect
Solution
Step 1:
L1={1,2,3} (3 points),L2={3,4,5,6,7} (5 points)L_1 = \left\{1,2,3\right\}\ \left(3\ \text{points}\right),\qquad L_2 = \left\{3,4,5,6,7\right\}\ \left(5\ \text{points}\right)
 overlapping in the single point 33. Apart from these, no three of the seven are collinear. Step 2: (1) Each pair determines a line, but the collinear sets are over-counted:
(72)−(32)−(52)+2=21−3−10+2=10\binom72-\binom32-\binom52+2 = 21-3-10+2 = 10
 The (32)\binom32 pairs inside L1L_1 all give one line, so subtract 33 and add back 11; likewise for L2L_2. (1) holds. Step 3: (2) Choose 33 points, discarding collinear triples:
(73)−(33)−(53)=35−1−10=24\binom73-\binom33-\binom53 = 35-1-10 = 24
 **(B) holds.** Step 4: (3) A quadrilateral needs four of the points with no three of them collinear. Let jj be the number chosen from {1,2}\left\{1,2\right\}, kk the number from {4,5,6,7}\left\{4,5,6,7\right\}, and e=1e = 1 if point 33 is chosen. Being on L1L_1 means being counted by j+ej+e; on L2L_2, by k+ek+e. We need
j+e≤2,k+e≤2,j+k+e=4j+e \le 2 ,\qquad k+e \le 2 ,\qquad j+k+e = 4
 ee is 00 or 11, so the two cases below are all of them. - e=1e = 1: then j≤1j \le 1 and k≤1k \le 1, so j+k≤2≠3j+k \le 2 \ne 3. Impossible, point 33 can never be a vertex. e=0e = 0: then j≤2j \le 2, k≤2k \le 2 and j+k=4j+k = 4, forcing j=k=2j = k = 2:
(22)(42)=1×6=6\binom22\binom42 = 1\times6 = 6
Total=6\text{Total} = 6
 (3) holds. Step 5: (4) A circle is determined by three non-collinear points, and two different triples give the same circle only when four or more of the points happen to be concyclic. The maximum possible count is therefore the number of non-collinear triples:
(73)−(33)−(53)=24\binom73-\binom33-\binom53 = 24
 (4) holds, equal to the triangle count for exactly that reason. (Direct enumeration over the seven points, with {1,2,3}\left\{1,2,3\right\} and {3,4,5,6,7}\left\{3,4,5,6,7\right\} as the only collinear sets, confirms 1010 lines, 2424 triangles and 66 quadrilaterals.) Answer: (1), (2), (3) and (4)
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