The equation of a circle is given as $(x-2)^2 + (y+3)^2 = 16$. (a) We need to find the center and radius of the circle. (b) We need to determine whether the point P(5,-3) lies inside, on, or outside the circle.

1. Problem Description

The equation of a circle is given as $(x-2)^2 + (y+3)^2 = 16$.

(a) We need to find the center and radius of the circle.

(b) We need to determine whether the point P(5,-3) lies inside, on, or outside the circle.

2. Solution Steps

(a) The general equation of a circle with center $(h,k)$ and radius $r$ is given by

$(x-h)^2 + (y-k)^2 = r^2$.

Comparing this with the given equation $(x-2)^2 + (y+3)^2 = 16$, we have:

$h=2$, $k=-3$, and $r^2 = 16$.

Therefore, the center of the circle is $(2, -3)$ and the radius is $r = \sqrt{16} = 4$.

(b) To determine if the point P(5,-3) lies inside, on, or outside the circle, we need to calculate the distance between the point P and the center of the circle (2,-3) and compare it to the radius of the circle.

The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

The distance between P(5,-3) and the center (2,-3) is

$d = \sqrt{(5-2)^2 + (-3-(-3))^2} = \sqrt{(3)^2 + (0)^2} = \sqrt{9+0} = \sqrt{9} = 3$.

Since the distance $d = 3$ is less than the radius $r = 4$, the point P(5,-3) lies inside the circle.

3. Final Answer

(a) The center of the circle is (2, -3) and the radius is

4. (b) The point P(5,-3) lies inside the circle.