COMEDK2022MathematicsCircle
S x^2+y^2+2 x+3 y+1=0 and S^ x^2+y^2+4 x+3 y+2=0 are two circles. The point (-3,-2) lies
Options
- Ainside S^ only
- Binside Sonly
- Cinside S and S^
- Doutside S and S^
Correct answer
A. inside S^ only
Step-by-step solution
Let S(x, y) = x^2 + y^2 + 2x + 3y + 1 and S'(x, y) = x^2 + y^2 + 4x + 3y + 2 . To determine the position of the point P(-3, -2) with respect to the circles, we evaluate the power of the point for each circle. For circle S : S(-3, -2) = (-3)^2 + (-2)^2 + 2(-3) + 3(-2) + 1 S(-3, -2) = 9 + 4 - 6 - 6 + 1 = 2 . Since S(-3, -2) > 0 , the point (-3, -2) lies outside circle S . For circle S' : S'(-3, -2) = (-3)^2 + (-2)^2 + 4(-3) + 3(-2) + 2 S'(-3, -2) = 9 + 4 - 12 - 6 + 2 = -3 . Since S'(-3, -2) Therefore, the point lies