Let circle S1: x² + y² - 4x - 6y + 12 = 0 and circle S2: (x-5)² + (y-6)² = r² (r 1). (A) If S1 and S2 are internally tangent, then (r-1)² is divisible by (B) If S1 and S2 are externally tangent, then r² + 2r + 3 is divisible by (C) If S1 and S2 cut orthogonally, then r² - 1 is divisible

Question

Let circle S1: x² + y² - 4x - 6y + 12 = 0 and circle S2: (x-5)² + (y-6)² = r² (r > 1). (A) If S1 and S2 are internally tangent, then (r-1)² is divisible by (B) If S1 and S2 are externally tangent, then r² + 2r + 3 is divisible by (C) If S1 and S2 cut orthogonally, then r² - 1 is divisible by (D) If S1 and S2 intersect such that the common chord is longest, then r² + 5 is divisible by Options for each: 3, 4, 5, 6

Accepted Answer

5. For each case, find r, compute the given expression, and identify which provided number it is divisible by.

Solution

Related JEE Advanced Maths questions