Equation of the Circle

The formula for a circle is:

R^2 = (x - cx)^2 + (y - cy)^2

Where R is the radius and (cx, cy) is the center of the circle. So for each question you need to know the center and the radius. In number 1, they give you the center, you need to find the radius, in 2, you need to figure out the center (the midpoint of the diameter) and the radius.

Given the point (cx, cy), what does c represent? Also, can you take a look at the other question concerning the diagonals of a parallelogram? Let me know if I did this right?

Why so many questions? I am traveling through a precalculus textbook chapter by chapter and often get stuck somewhere along the way. Keep in mind that I do answer most of the chapter questions correctly.

Cx, and Cy are the x and y coordinates of the center of the circle.

If you set both Cx and Cy to 0, you get a circle with the center at (0,0).

If you do this, do you see what happens to the equation?

Let cx = 0 and cy = 0.

R^2 = (x - 0)^2 + (y - 0)^2

R^2 = x^2 + y^2

This is the general form for the equation of a circle centered at the origin.
In what way does this lead to the answer?

cx is just the X coordinate of the center, cy is the Y coordinate, so if the center is (−2, −6), the formula is going to look like R^2 = (x+2)^2 + (y+6)^2. You need to find R^2 to complete the formula. If you just plug in the point given, you should get the answer.

Your parallelogram problem looked fine.

Engineer is being a little silly because I hurt his feelings on a political thread. That is why on math threads he is repeating answers that I have already given you.

On a math thread, it is a ridiculous.

Because is cx and cy represent the x and y coordinates of the circle... than setting cx,xy to (0,0) is a circle (centered at the origin). See how it makes sense?

Inserting different interesting values to see if it makes sense is a good way to to understand a mathematical relation.

How does this in any way answer questions 1 and 2?

1. Center: (−2, −6);
Solution point: (1, −10)

R^2 = (x -(-2))^2 + (y - (-6))^2

R^2 = (x + 2)^2 + (y + 6)^2

To find R, I must use the distance formula.

Yes?

R = sqrt{(-2 - 1)^2 + (-6 -(-10))^2}

R = sqrt{(-3)^2 + (-6 + 10)^2}

R = sqrt{9 + 16}

R = sqrt{25}

R = 5

If R = 5, then R^2 = 25.

The equation is as follows:

25 = (x + 2)^2 + (y + 6)^2

Is this right?


2. Endpoints of a diameter:
(11, −5), (3, 15).

Let M = midpoint of diameter.

M = [(11 + 3)/2, (-5 + 15)/2]

M = (14/2, 10/2)

M = (7, 5)

I see that the equation is not centered at the origin.

R = sqrt{(7 - 3)^2 + (5 - 15)^2}

R = sqrt{(4)^2 + (-10)^2}

R = sqrt{16 + 100}

R = sqrt{116}

If R = sqrt{116}, then R^2 = 116.

The equation is as follows:

116 = (x - 7)^2 + (y - 5)^2

Is this right?

Your methodology is right, but in the first one you forgot to square 16.

What do you mean forgot to square 16?


R = sqrt{(-2 - 1)^2 + (-6 -(-10))^2}

R = sqrt{(-3)^2 + (-6 + 10)^2}

R = sqrt{9 + 16}

R = sqrt{25}

R = 5

If R = 5, then R^2 = 25.

The equation is as follows:

25 = (x + 2)^2 + (y + 6)^2

Yes?

My mistake, you are correct, nice job.

Thank you. Look for questions as I travel through my precalculus textbook.