If the circles (x+1)²+(y+2)²=r² and x²+y²-4 x-4 y+4=0 intersect at exactly two distinct points, then

Solution

Maths sir · 4 steps

Find both centres and radii, then trap the distance between centres between |r-2| and r+2.

1. Read off both circles. The first circle has centre C₁(-1,-2) and radius r. The second, x²+y²-4x-4y+4=0, has centre C₂(2,2) and radius √(4+4-4)=2.
2. Distance between centres. C₁C₂ = √((2+1)²+(2+2)²) = √(9+16) = 5.

   |r₁ - r₂| < C₁C₂ < r₁ + r₂ ⇒ |r-2| < 5 < r+2

3. Solve both inequalities. From 5 < r+2 we get r > 3, so the circles are not too far apart. From |r-2| < 5 we get -3 < r < 7, so neither circle lies inside the other.
4. So. Both conditions together give 3 < r < 7, which is option (b).

The idea to keep. Two circles cut at two points exactly when the distance between centres is more than the difference of the radii and less than their sum. Always use both inequalities.

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