Question

If a variable line, \(3x + 4y - \lambda = 0\) is such that the two circles \(x^2 + y^2 - 2x - 2y + 1 = 0\) and \(x^2 + y^2 - 18x - 2y + 78 = 0\) are on its opposite sides, then the set of all values of \(\lambda\) is the interval :

• A. \((23, 31)\)
• B. \((2, 17)\)
• C. \([13, 23]\)
• D. \([12, 21]\)

Hint:

Find the centres and radii; the line must be at distance at least the radius from each centre, with the centres on opposite sides.

Answer 1

The Correct Option is D

Circle 1: centre \((1,1)\), radius \(\sqrt{1+1-1} = 1\). Circle 2: centre \((9,1)\), radius \(\sqrt{81+1-78} = 2\).
Value of \(3x+4y-\lambda\) at the centres: \(7-\lambda\) and \(31-\lambda\). For opposite sides these must have opposite signs: \((7-\lambda)(31-\lambda) \lt 0 \Rightarrow 7 \lt \lambda \lt 31\).
Circle 1 lies wholly on its side: \(\frac{|7-\lambda|}{5} \ge 1 \Rightarrow \lambda - 7 \ge 5 \Rightarrow \lambda \ge 12\).
Circle 2 lies wholly on its side: \(\frac{|31-\lambda|}{5} \ge 2 \Rightarrow 31 - \lambda \ge 10 \Rightarrow \lambda \le 21\).
Hence \(\lambda \in [12, 21]\).
Answer: (4) \([12, 21]\)