Question

1. Find the position of the points (1, 2) & (6, 0) w.r.t. the circle $x^2 + y^2 - 4x + 2y - 11 = 0$

Answer 1

Point (1, 2) is inside the circle, and point (6, 0) is outside the circle.

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 $S = x^2 + y^2 - 4x + 2y - 11 = 0$.

For point (1, 2): $S(1, 2) = (1)^2 + (2)^2 - 4(1) + 2(2) - 11 = 1 + 4 - 4 + 4 - 11 = -6$. Since $S(1, 2) < 0$, the point (1, 2) is inside the circle.

For point (6, 0): $S(6, 0) = (6)^2 + (0)^2 - 4(6) + 2(0) - 11 = 36 + 0 - 24 + 0 - 11 = 1$. Since $S(6, 0) > 0$, the point (6, 0) is outside the circle.