Equation of a Circle

Let's start by recalling the standard equation of a circle. (x- h)^2+(y- k)^2= r^2 We will rewrite the given equation to match this form. In this case, we will need to complete the square.

We can now simplify this a little. (x- 2)^2+(y- 0)^2= 3^2 ⇓ (x-2)^2+y^2=9

Let's have a look at the standard equation of a circle. (x- h)^2+(y- k)^2=r^2 Here, the center is the point ( h, k). We can use this to identify the center of our circle. (x- 2)^2+(y- 0)^2=3^2 The circle has its center in ( 2, 0).

The radius is written as r in the standard equation of a circle. (x-h)^2+(y-k)^2= r^2 Let's compare this with the expression for our circle. (x-2)^2+(y-0)^2= 3^2 We can see that our circle has a radius of 3.

In Part B, we found that the circle has its center in ( 2, 0). In Part C, we learned that the circle has a radius of 3. By using this information we can draw its graph.

We can now conclude that graph iv represents the circle.

Let's recall the standard equation of a circle. (x- h)^2+(y- k)^2= r^2 We will rewrite the given equation to match this form. In this case, we will need to complete the square twice — once for each variable.

We can now simplify this a little. (x-( - 2))^2+(y-( - 6))^2= 5^2 ⇓ (x+2)^2+(y+6)^2=25

Let's once more look into the standard equation of a circle. (x- h)^2+(y- k)^2=r^2 Here, the center is the point ( h, k). We can use this to identify the center of our circle. (x-( - 2))^2+(y-( - 6))^2=5^2 The circle has its center in ( - 2, - 6).

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 for our circle. (x-(- 2))^2+(y-(- 6))^2= 5^2 We can see that our circle has a radius of 5.

In Part B we found that the circle has its center in ( - 2, - 6). In Part C we learned that the circle has a radius of 5. By using this information we can draw its graph.

We can now identify that graph i represents the circle.

We will first find the circle's radius. When we know the center and the radius we can write the circle's equation. Finally, we will use the equation to find the circle's second x-intercept.

Finding the Radius

Let's recall the Distance Formula. d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) We find the radius of our circle by substituting the center and the x-intercept into this equation, then evaluating the right-hand side.

The circle has a radius of sqrt(18) 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 exercise tells us that the center of the circle is ( - 2, 3) and we have calculated its radius and found that it is sqrt(18) units. Let's use this knowledge to write the equation. (x-( - 2))^2+(y- 3)^2=( sqrt(18))^2 ⇕ (x+2)^2+(y-3)^2=18

Finding the x-intercept

We find the circle's x-intercepts by substituting 0 for y into the equation and solving for x.

The circle has two x-intercepts. One of them (- 5,0) was given to us in the exercise. The other is at (1,0).

We will first write the equation of the circle. Then we will use this equation to find the circle's y-intercept(s).

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 ( 3, 1) and it has a radius of 4. Let's use this to write the equation. (x- 3)^2+(y- 1)^2= 4^2 ⇕ (x-3)^2+(y-1)^2=16

Finding the y-intercepts

We find the circle's y-intercept(s) by substituting 0 for x into the equation, then solve for y.