The problem provides the equation of a circle, $(x-2)^2 + (y+3)^2 = 16$. Part (a) asks for the center and radius of the circle. Part (b) asks whether the point $P(5,-3)$ lies inside, on, or outside the circle.

1. Problem Description

The problem provides the equation of a circle, $(x-2)^2 + (y+3)^2 = 16$.

Part (a) asks for the center and radius of the circle.

Part (b) asks whether the point $P(5,-3)$ lies inside, on, or outside the circle.

2. Solution Steps

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

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

Comparing this to the given equation $(x-2)^2 + (y+3)^2 = 16$, we can identify the center and radius. We have $h = 2$, $k = -3$, and $r^2 = 16$. Taking the square root of $r^2$ gives $r = 4$.

So the center of the circle is $(2, -3)$ and the radius is $4$.

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

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

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

In our case, $(x_1, y_1) = (2, -3)$ (center of the circle) and $(x_2, y_2) = (5, -3)$ (point $P$). Therefore, the distance $d$ 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) Center: $(2, -3)$, Radius: $4$

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