Question

S $ \equiv x^2 + y^2 - 2x - 4y - 4 = 0 $ and $ S' \equiv x^2 + y^2 - 4x - 2y - 16 = 0 $ are two circles. The point $ (-2, -1) $ lies

• A. inside \( S' \) only
• B. inside \( S \) only
• C. inside \( S \) and \( S' \)
• D. outside \( S \) and \( S' \)

Hint:

When determining whether a point lies inside or outside a circle, substitute the point's coordinates into the equation of the circle. If the result is less than 0, the point is inside; if greater than 0, the point is outside.

Answer 1

The Correct Option is A

We are given two circles with the equations \( S \) and \( S' \). To determine whether the point \( (-2, -1) \) lies inside or outside the circles, we need to substitute the point coordinates into the equations of the circles. 
Circle \( S: x^2 + y^2 - 2x - 4y - 4 = 0 \) Substitute \( x = -2 \) and \( y = -1 \) into the equation: \[ (-2)^2 + (-1)^2 - 2(-2) - 4(-1) - 4 = 4 + 1 + 4 + 4 - 4 = 9. \] Since \( 9>0 \), the point lies outside circle \( S \). Circle \( S': x^2 + y^2 - 4x - 2y - 16 = 0 \) Substitute \( x = -2 \) and \( y = -1 \) into the equation: \[ (-2)^2 + (-1)^2 - 4(-2) - 2(-1) - 16 = 4 + 1 + 8 + 2 - 16 = -1. \] Since \( -1<0 \), the point lies inside circle \( S' \). 
Thus, the point \( (-2, -1) \) lies inside circle \( S' \) only.