If the circles x² + y² + 6x + 8y + 16 = 0 and x² + y² + 2( 3 - √3 )x + 2( 4 - √6 )y = k + 6√3 + 8√6 , k > 0, touch internally at the point P(α ,β ), then ( α + √3 )² + ( β + √6 )² is equal to ________________.

Solution

Maths sir · 5 steps

Find both centres and radii first; the contact point sits on the line joining the centres.

1. First circle. Centre C₁ = (-3, -4) and radius r₁ = √(9 + 16 - 16) = 3.
2. Second circle. Centre C₂ = (√3 - 3, √6 - 4). For the radius, r₂² = (3 - √3)² + (4 - √6)² + k + 6√3 + 8√6. The surd terms cancel, which leaves r₂² = 12 + 22 + k = 34 + k.
3. Use internal touching. C₁ - C₂ = (-√3, -√6), so C₁C₂ = √(3 + 6) = 3. Circles that touch internally need C₁C₂ = r₂ - r₁, which gives r₂ = 6. Then 34 + k = 36, so k = 2 > 0, as required.
4. Locate the point. P lies on the line through C₂ and C₁, beyond C₁, at distance r₂ = 6 from C₂. The unit vector from C₂ towards C₁ is 1/3(-√3, -√6).

   P = C₂ + 6 · 1/3(-√3, -√6) = (-3 - √3, -4 - √6)

5. So. α + √3 = -3 and β + √6 = -4. The required value is 9 + 16 = 25.

The idea to keep. When two circles touch, the point of contact always lies on the line of centres. For internal touching use d = |r₁ - r₂|, then step out from the bigger circle's centre by its radius.

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