Question

Add the following method to your Circle class, which given another point's x and y coordinates, this method returns whether or not that point is within the bounds of the Circle or not: public bool Contains(float px, float py).To help you accomplish this, on a piece of paper, draw a circle with a center point and radius, then draw two other points, one inside the circle and one outside. How could the computer compute which point is inside the circle and which point is outside the circle? Think about the radius of the circle from its center point compared to the distance between the point and the center of the circle... For this Test, when you have finished adding the above requested information to your Circle class, you will then make an instance object of type Circle, calling its overloaded constructor passing the variables given to you in this test and then you will return your class instance object from this test.

Answer 1

To determine if a point is within the bounds of a circle, compare the distance between the point and the center of the circle to the radius. If the distance is less than or equal to the radius, the point is inside; otherwise, it is outside. You can implement this in the Circle class by calculating the distance using the distance formula and comparing it to the radius.

Step-by-step explanation:

The method to determine if a point is within the bounds of a circle involves comparing the distance between the point and the center of the circle to the radius of the circle. If the distance is less than or equal to the radius, the point is inside the circle; otherwise, it is outside in the Circle class, you can implement the Contains method by calculating the distance between the given point (px, py) and the center of the circle using the distance formula: distance = sqrt((px - center_x)^2 + (py - center_y)^2). Then, you can compare this distance to the radius of the circle.

To demonstrate this, let's say the circle has a center point (5, 5) and a radius of 3. We can check if another point, let's say (4, 4), is within the bounds of the circle. The distance between (4, 4) and (5, 5) is sqrt((4 - 5)^2 + (4 - 5)^2) = sqrt(2). Since sqrt(2) < 3, the point (4, 4) is inside the circle.