Question

Consider the points P(2, 1); Q(0, 0); R(4, -3) and the circle $S : x^2 + y^2 - 5x + 2y - 5 = 0$

• A. exactly one point lies outside $S$
• B. exactly two points lie outside $S$
• C. all the three points lie outside $S$
• D. none of the point lies outside $S$

Answer 1

Answer: (D)

none of the point lies outside $S$

Answer 2

To determine the position of a point $(x_1, y_1)$ with respect to a circle $S = x^2 + y^2 + 2gx + 2fy + c = 0$, we evaluate $S(x_1, y_1)$:

• If $S(x_1, y_1) > 0$, the point is outside the circle.
• If $S(x_1, y_1) = 0$, the point is on the circle.
• If $S(x_1, y_1) < 0$, the point is inside the circle.

The given circle equation is $S : x^2 + y^2 - 5x + 2y - 5 = 0$. The given points are P(2, 1), Q(0, 0), and R(4, -3).

Let's evaluate $S(x, y)$ for each point:

1. For point P(2, 1): $S(2, 1) = (2)^2 + (1)^2 - 5(2) + 2(1) - 5$ $S(2, 1) = 4 + 1 - 10 + 2 - 5$ $S(2, 1) = -8$ Since $S(2, 1) < 0$, point P is inside the circle.

2. For point Q(0, 0): $S(0, 0) = (0)^2 + (0)^2 - 5(0) + 2(0) - 5$ $S(0, 0) = -5$ Since $S(0, 0) < 0$, point Q is inside the circle.

3. For point R(4, -3): $S(4, -3) = (4)^2 + (-3)^2 - 5(4) + 2(-3) - 5$ $S(4, -3) = 16 + 9 - 20 - 6 - 5$ $S(4, -3) = -6$ Since $S(4, -3) < 0$, point R is inside the circle.

All three points P, Q, and R are inside the circle S. Therefore, none of the points lie outside S.