Equation of a Circle

The center is in the middle of the circle. Let's use the diagram to identify its coordinates.

The circle has its center in the point (4,3).

The radius of a circle is the distance between its center and any point on the circle's circumference. We found the center in Part A. Let's now identify a point on the circle that we can use to calculate the radius.

Since the endpoints of the segment have the same y-coordinate we can use the Ruler Postulate to find its length. r=|7-4|=3 The circle has a radius of 3.

Let's recall the standard equation of a circle with center in the point ( h, k) and a radius of r. (x- h)^2+(y- k)^2= r^2 In Part A, we found that the circle has its center in ( 4, 3), and in Part B, we found that the radius of the circle is 3. We find the equation of the circle by substituting these into the expression. (x- 4)^2+(y- 3)^2= 3^2

Let's start by recalling the standard equation of a circle. (x- h)^2+(y- k)^2= r^2 The center is the point ( h, k) in this equation. Let's rewrite the given equation to match this form.

We can now use this equation to identify the center of our circle. (x- 0)^2+(y-( - 1))^2=2^2 The circle has its center in ( 0, - 1).

The radius is denoted r in the standard equation of a circle. (x-h)^2+(y-k)^2= r^2 Let's compare this with the expression we found in Part A. (x-0)^2+(y-(- 1))^2= 2^2 We can see that our circle has a radius of 2.

In Part A we found that the circle has its center in ( 0, - 1). In Part B we learned that the circle has a radius of 2. By using this information we can draw its graph.

We can now conclude that the graph iii represents the circle.

Let's recall the standard equation of a circle. (x- h)^2+(y- k)^2= r^2 In this form, ( h, k) is the center of the circle and r is its radius. The exercise tells us that the center of the circle is ( 9, 0) and its radius is 5. Let's use this to write the equation. (x- 9)^2+(y- 0)^2= 5^2 ⇕ (x-9)^2+y^2=25

Let's start by recalling the standard equation of a circle. (x- h)^2+(y- k)^2= r^2 In this form, ( h, k) is the center of the circle and r is its radius. We are told that the center of the circle is ( 1, -1) and its radius is sqrt(5). Now we have all the information we need to write the equation. (x- 1)^2+(y-( -1))^2=( sqrt(5))^2 ⇕ (x-1)^2+(y+1)^2=5

Let's recall the Distance Formula. d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) Using this formula, we can find the distance d between two points (x_1,y_1) and (x_2,y_2).

The circle has a radius of 5.

Let's recall the standard equation of a circle. (x- h)^2+(y- k)^2= r^2 In this form, ( h, k) is the center of the circle and r is its radius. The exercise tells us that the center of the circle is ( 3, - 4) and its radius is 5. Let's use this to write the equation. (x- 3)^2+(y-( - 4))^2= 5^2 ⇕ (x-3)^2+(y+4)^2=25

We have been given that the circle has its center in (3,- 4) and contains the point (0,- 8). Let's mark these points in a diagram and use them to graph the circle.

We can now identify that graph i represents the circle.

We will first find the center and the radius of the circle. Then we will use these to write its equation.

Finding Center and Radius

Let's identify the coordinates of the center and the coordinates of one point on the circle.

Choosing a point that is on the same y-coordinate as the center, we can find the radius by using the Ruler Postulate. r=|2-7|= 5 The circle has a radius of 5 units.

Writing the Equation

Let's recall the standard equation of a circle. (x- h)^2+(y- k)^2= r^2 In this form, ( h, k) is the center of the circle and r is its radius. The center of the circle is ( 2, 3) and its radius is 5. Let's substitute this into the standard equation. (x- 2)^2+(y- 3)^2= 5^2 ⇕ (x-2)^2+(y-3)^2=25

Let's recall the standard equation of a circle. (x- h)^2+(y- k)^2= r^2 In this form, ( h, k) is the center of the circle and r is its radius. In the exercise we are told that center of the circle is at the origin ( 0, 0). Let's substitute this into the equation. (x- 0)^2+(y- 0)^2= r^2 ⇕ x^2+y^2= r^2 We know that the diameter of the Buzludzha monument is 70 meters. The radius must be half that which is r= 35 meters. Let's substitute the radius into the equation. x^2+y^2= 35^2 ⇕ x^2+y^2=1225