JEE MainMathematicsCircle
If the circles x^2 + y^2 = 49 and x^2 + y^2 - 2kx - 6y + k^2 + 5 = 0 intersect at exactly two distinct points, then the range of values of k is
Options
- Ak (-6 2 , 6 2 )
- Bk (- , -4) (4, )
- Ck (-4, 4)
- Dk (-6 2 , -4) (4, 6 2 )
Correct answer
D. k (-6 2 , -4) (4, 6 2 )
Step-by-step solution
Let the given circles be S₁ x^2 + y^2 = 49 and S₂ x^2 + y^2 - 2kx - 6y + k^2 + 5 = 0 . The center and radius of S₁ are C₁(0, 0) and r₁ = 7 . The center and radius of S₂ are C₂(k, 3) and r₂ = k^2 + 9 - (k^2 + 5) = 4 = 2 . For two circles to intersect at exactly two distinct points, the distance between their centers d must satisfy: |r₁ - r₂| The distance between the centers is d = k^2 + 3^2 = k^2 + 9 . Substituting the values into the condition: |7 - 2| 5 Squaring all parts of the inequality: 25 16 This gives two in