COMEDK2022MathematicsCircle
If two circles (x-1)^2+(y-3)^2=r^2 and x^2+y^2-8 x+2 y+8=0 intersect in two distinct points, then
Options
- A2 < r < 8
- Br < 2
- Cr=2
- Dr>2
Correct answer
A. 2 < r < 8
Step-by-step solution
The first circle is (x-1)^2 + (y-3)^2 = r^2 , which has center C₁ = (1, 3) and radius r₁ = r . The second circle is x^2 + y^2 - 8x + 2y + 8 = 0 . Completing the square: (x^2 - 8x + 16) + (y^2 + 2y + 1) = -8 + 16 + 1 , which simplifies to (x-4)^2 + (y+1)^2 = 9 . The second circle has center C₂ = (4, -1) and radius r₂ = 3 . The distance between the centers C₁ and C₂ is d = (4-1)^2 + (-1-3)^2 = 3^2 + (-4)^2 = 9 + 16 = 5 . Two circles intersect in two distinct points if and only if |r₁ - r₂| Substituting the values: |r