Let the equation x² + y² + px + (1 - p)y + 5 = 0 represent circles of varying radius r ∈ (0, 5]. Then the number of elements in the set S = q : q = p² and q is an integer is __________.

Solution

Maths sir · 3 steps

Write r² in terms of p, trap it in (0, 25], then see which squares p² can reach.

1. Radius squared. For x² + y² + 2gx + 2fy + c = 0, r² = g² + f² - c. Here g = p/2, f = (1-p)/2 and c = 5.

   r² = (p² + (1-p)²)/4 - 5 = (2p² - 2p + 1)/4 - 5

2. Apply the range. The condition 0 < r² ≤ 25 becomes 20 < 2p² - 2p + 1 ≤ 120. The left inequality 2p² - 2p - 19 > 0 gives p < (1-√39)/2 or p > (1+√39)/2. The right inequality 2p² - 2p - 119 ≤ 0 gives (1-√239)/2 ≤ p ≤ (1+√239)/2.

   Branch of p	Interval for p	Interval for q = p²
   negative	[-7.23, -2.62)	(10 - √39/2, 60 - √239/2] ≈ (6.88, 52.27]
   positive	(3.62, 8.23]	(10 + √39/2, 60 + √239/2] ≈ (13.12, 67.73]

3. So. Together the two ranges cover (6.88, 67.73], so the integer values of q run from 7 to 67. That is 67 - 7 + 1 = 61 elements.

The common slip: 54. Using only the positive values of p makes q run from 14 to 67, which gives 54. The negative branch adds the integers 7 to 13.

The idea to keep. When a question asks about p², check both signs of p. The negative branch often pushes the lower end of the range of p² further down than the positive branch does.

Answer from the NTA final answer key. Solution by PrepWiser.