Question

Let L1:4x−3y+13=0, L2:4x−3y=37, L3:3x+4y=34 are three lines in xy plane and L4:(1+λ)x+(1−λ)y=24 is a variable line. P(a,b) is centre of circle which touches lines L1, L2 and L3. Maximum value of a+b is-

Answer 1

17

Answer 2

1. The circle touches L₁: 4x – 3y + 13 = 0 and L₂: 4x – 3y – 37 = 0. Since these lines are parallel, the distance between them is

   Distance = |37 – (–13)|/5 = 50/5 = 10.

   Hence, the radius r = 10/2 = 5.

2. For the center P(a, b) of the circle, we have:

   Distance from P to L₁: |4a – 3b + 13|/5 = 5 → |4a – 3b + 13| = 25

   Distance from P to L₂: |4a – 3b – 37|/5 = 5 → |4a – 3b – 37| = 25

   Let k = 4a – 3b. Then:

   |k + 13| = 25 and |k – 37| = 25.

   Solving, k + 13 = 25 gives k = 12, and indeed |12 – 37| = 25.

   Thus,

   4a – 3b = 12 (1)

3. The circle also touches L₃: 3x + 4y – 34 = 0, so:

   |3a + 4b – 34|/5 = 5 → |3a + 4b – 34| = 25

   This gives two cases:

   Case I: 3a + 4b – 34 = 25 → 3a + 4b = 59

   Case II: 3a + 4b – 34 = –25 → 3a + 4b = 9

4. Solving Case I:

   From (1), 4a – 3b = 12 → a = (12 + 3b)/4

   Substitute into 3a + 4b = 59:

   3[(12 + 3b)/4] + 4b = 59

   (36 + 9b)/4 + 4b = 59

   Multiply through by 4: 36 + 9b + 16b = 236

   25b = 200 → b = 8

   Then, a = (12 + 3×8)/4 = (12 + 24)/4 = 9

   So, a + b = 9 + 8 = 17

5. Solving Case II:

   Using a = (12 + 3b)/4 in 3a + 4b = 9:

   3[(12 + 3b)/4] + 4b = 9

   (36 + 9b)/4 + 4b = 9

   Multiply through by 4: 36 + 9b + 16b = 36

   25b = 0 → b = 0

   Then, a = (12)/4 = 3

   So, a + b = 3 + 0 = 3

6. Conclusion: The maximum possible value of a + b is 17 (from Case I).