The number of common tangents, to the circles
x²+y²-18 x-15 y+131=0
and x²+y²-6 x-6 y-7=0, is :

Solution

Maths sir · 4 steps

Find both centres and radii, then compare the distance between centres with the sum of the radii.

1. First circle. For x²+y²+2gx+2fy+c=0 the centre is (-g,-f) and r² = g²+f²-c. Here the centre is (9, 15/2) and r₁² = 81 + 56.25 - 131 = 6.25, so r₁ = 2.5.
2. Second circle. The centre is (3, 3) and r₂² = 9 + 9 + 7 = 25, so r₂ = 5.

   C₁C₂ = √((9-3)² + (7.5-3)²) = √(36 + 20.25) = √56.25 = 7.5

3. Compare. r₁ + r₂ = 2.5 + 5 = 7.5, which equals C₁C₂. So the circles touch each other externally.
4. So. Two circles touching externally have two direct common tangents and one transverse tangent at the point of contact. That makes 3 common tangents, option (c).

The idea to keep. Count common tangents by comparing d = C₁C₂ with the radii: d > r₁+r₂ gives 4, d = r₁+r₂ gives 3, |r₁-r₂| < d < r₁+r₂ gives 2, d = |r₁-r₂| gives 1, and a smaller d gives 0.

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