Suppose a circle is centered at the vertex (0,0) and has a radius of length 4.


For which points on the circle is the x-coordinate equal to 1? Your first answer should have a positive y-value.

For which points on the circle is the x-coordinate equal to 3? Your first answer should have a positive y-value.

What is the equation of the graph of the circle?

Well, when the x-coordinate is equal to 1, we just need to find the corresponding y-coordinate on the circle. Since the circle is centered at (0,0), we can use the Pythagorean theorem to find the y-coordinate.

The equation of the circle is x^2 + y^2 = r^2, where r is the radius. So, substituting r = 4 into the equation, we have x^2 + y^2 = 16.

When x = 1, we can solve for y:

1^2 + y^2 = 16
1 + y^2 = 16
y^2 = 15
y = √15 or approximately 3.87

Therefore, when the x-coordinate is equal to 1, the corresponding point on the circle is (1, √15).

Now, let's move on to when the x-coordinate is equal to 3. Similarly, we can solve for the y-coordinate:

3^2 + y^2 = 16
9 + y^2 = 16
y^2 = 7
y = √7 or approximately 2.65

So, when the x-coordinate is equal to 3, the corresponding point on the circle is (3, √7).

The equation of the graph of the circle is x^2 + y^2 = 16. It's like a mathematical barrier around the origin, ensuring that no point gets too close without proper identification.

To find the points on the circle where the x-coordinate is equal to 1, we can substitute x = 1 into the equation of the circle and solve for y.

Step 1: Start with the equation of a circle in the form (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r is the radius.

In this case, the center of the circle is at (0, 0), so the equation becomes x^2 + y^2 = 4^2.

Step 2: Substitute x = 1 into the equation.

1^2 + y^2 = 4^2

1 + y^2 = 16

Step 3: Solve for y.

y^2 = 16 - 1

y^2 = 15

Taking the square root of both sides, we get:

y = ± √15

Since we need a positive y-value, the first answer where the x-coordinate is 1 and y is positive is (1, √15).

For the points on the circle where the x-coordinate is equal to 3, we repeat the same steps.

Step 1: Start with the equation of the circle: x^2 + y^2 = 4^2.

Step 2: Substitute x = 3 into the equation.

3^2 + y^2 = 4^2

9 + y^2 = 16

Step 3: Solve for y.

y^2 = 16 - 9

y^2 = 7

Taking the square root of both sides, we get:

y = ± √7

Since we need a positive y-value, the first answer where the x-coordinate is 3 and y is positive is (3, √7).

Therefore, the points on the circle where the x-coordinate is equal to 1 are (1, √15) and (1, -√15), and the points where the x-coordinate is equal to 3 are (3, √7) and (3, -√7).

The equation of the graph of the circle is x^2 + y^2 = 16.

To find the points on the circle where the x-coordinate is equal to 1, we can substitute x = 1 into the equation of the circle and solve for y.

The equation of a circle centered at (h, k) with a radius of r is given by (x - h)^2 + (y - k)^2 = r^2.

In this case, the circle is centered at (0, 0) and has a radius of 4, so the equation of the circle is x^2 + y^2 = 16.

Substituting x = 1 into the equation, we get 1^2 + y^2 = 16.

Simplifying the equation, we have 1 + y^2 = 16.

Subtracting 1 from both sides, we get y^2 = 15.

Taking the square root of both sides, we have y = ±√15.

Since we are looking for the positive value of y, the point on the circle where the x-coordinate is equal to 1 is (1, √15).

To find the points on the circle where the x-coordinate is equal to 3, we can follow the same steps.

Substituting x = 3 into the equation of the circle, we get 3^2 + y^2 = 16.

Simplifying the equation, we have 9 + y^2 = 16.

Subtracting 9 from both sides, we get y^2 = 7.

Taking the square root of both sides, we have y = ±√7.

Since we are looking for the positive value of y, the point on the circle where the x-coordinate is equal to 3 is (3, √7).

Therefore, the points on the circle where the x-coordinate is equal to 1 are (1, √15) and (1, -√15), and the points on the circle where the x-coordinate is equal to 3 are (3, √7) and (3, -√7).

The equation of the graph of the circle is x^2 + y^2 = 16, which represents all points (x, y) that are 4 units away from the origin (0, 0) in any direction.

I will start with the last part:

equation of the circle is
x^2 + y^2 = 16

so if x = 1 ....
1 + y^2 = 16
y^2 = 15
y = √15 , actually ± √15 , but you wanted the positive

repeat my steps for x = 3