JEE MainMathematicsCircle
A circle S has its center on the x-axis and passes through the origin. If it intersects the circle x^2 + y^2 - 8x - 6y + 21 = 0 at exactly two distinct points, then the number of possible integral values for the x-coordinate of the center of S is
Options
- A4
- B5
- C3
- D6
Correct answer
A. 4
Step-by-step solution
Let the center of circle S be (a, 0) . Since it passes through the origin (0,0) , its radius is r₁ = (a-0)^2 + (0-0)^2 = |a| . For the given fixed circle x^2 + y^2 - 8x - 6y + 21 = 0 , the center is C₂(4, 3) and its radius is r₂ = 4^2 + 3^2 - 21 = 25 - 21 = 2 . The distance between the centers is d = (a - 4)^2 + (0 - 3)^2 = (a - 4)^2 + 9 . For the two circles to intersect at exactly two distinct points, we must have: |r₁ - r₂| ||a| - 2| Case 1: a 0 Then |a| = a . The inequality becomes: |a - 2| Squaring all parts (