a First, let's consider a circle with center at the the origin, point (0,0), that contains the point (0,3).
We want to determine if the point (1, sqrt(5)) lies on that circle. Notice that the distance between the center of our circle, point (0,0), and point (0,3) is 3 units. So, the radius of our circle is 3 units.
Therefore, our circle is the set of all points on a plane that are in distance 3 from point (0,0). To find if point (1,sqrt(5)) lies on our circle, we need to check if its distance to (0,0) is 3. We can do so using the Distance Formula.
d = sqrt((x_2-x_1)^2+(y_2-y_1)^2)
The above formula gives us the distance between points (x_1,y_1) and (x_2,y_2). Let's substitute (0,0) for (x_1,y_1) and (1,sqrt(5)) for (x_2,y_2) in this formula to find the distance between (0,0) and (1,sqrt(5)).
Now, since the distance between (0,0) and (1,sqrt(5)) is about 2.45, it is not equal to 3. For this reason, point (1,sqrt(5)) does not lie on our circle.
b In Part A, we found that our circle is the set of all point that are in distance of 3 units from point (0,0). Now we want to find at least one value of x so that point (x,sqrt(5)) lies on our circle. To do so, we can use the Distance Formula to write an equation for x.
d = sqrt((x_2-x_1)^2+(y_2-y_1)^2)
The above formula tells us the distance from (x_1,y_1) to (x_2,y_2). We can substitute (0,0) for (x_1,y_1), (x,sqrt(5)) for (x_2,y_2), and 3 for the distance d to write an equation for x.
d = sqrt((x_2-x_1)^2+(y_2-y_1)^2)
⇓
3 = sqrt(( x - 0)^2 + ( sqrt(5) - 0)^2)
Now, let's solve it!
Now, let's notice that x=2 and x=-2 both satisfy our equation.
x^2 = 4 ⇔ x = 2 or x=-2
c We want to name three other point that lie on our circle. To do so, we will first find the standard equation of our circle. First, let's recall the general form of a standard equation of a circle.
(x-x_0)^2+(y-y_0)^2 = r^2
Here, (x_0, y_0) is the center of the circle and r is its radius. The center of our circle is the point ( 0, 0) and in Part A we found that the radius r is 3 units. Let's use these values to find write the standard equation of our circle.
(x-x_0)^2+(y-y_0)^2 = r^2
⇓
(x- 0)^2 + (y- 0)^2 = 3^2Let's simplify this equation!
Now, we will find the points of our circle that lie on the axes. First, let's substitute 0 for x in the above equation and find the values of y that satisfy the equation.
Notice that y=3 and y=-3 both satisfy the resulting equation.
y^2 = 9 ⇔ y = 3 or y=-3
Therefore, points (0,3) and (0,-3) lie on our circle. We already knew that this is the case for the point (0,3), so (0,-3) is the only new point. Now, let's substitute 0 for y in the equation of our circle.
Notice that x=3 and x=-3 both satisfy the resulting equation.
x^2 = 9 ⇔ x = 3 or x=-3
Therefore, points (3,0) and (-3,0) lie on our circle. Along with point (0,-3), we have three new points on our circle.
(0,-3), (3,0), (-3,0)
