Math  /  Geometry

QuestionWrite the standard form of the equation of the circle with its center at ( 7,0-7,0 ), and a radrus of 3
What is the equation of the circle in standard form? \square (Simplify your answer.)

Studdy Solution

STEP 1

What is this asking? We need to find the equation of a circle with a given center and radius, and write it in the standard form. Watch out! Don't mix up the signs of the coordinates of the center in the equation!

STEP 2

1. Recall the standard form
2. Plug in the values

STEP 3

Alright, let's **kick things off** by remembering what the standard form of a circle's equation looks like.
It's (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the **center** of the circle, and rr is its **radius**.
This formula beautifully captures the essence of a circle: all the points (x,y)(x, y) that are a fixed distance rr from the center (h,k)(h, k).

STEP 4

Now, let's **plug in** our given values.
We know the center is at (7,0)(-7, 0), so h=7h = -7 and k=0k = 0.
The radius is 33, so r=3r = 3.
Substituting these values into our standard form equation, we get (x(7))2+(y0)2=32(x - (-7))^2 + (y - 0)^2 = 3^2.

STEP 5

Let's **simplify** this a bit.
Subtracting a negative is the same as adding, so (x(7))(x - (-7)) becomes (x+7)(x + 7).
Also, (y0)(y - 0) is just yy, and 323^2 is 99.
So, our equation becomes (x+7)2+y2=9(x + 7)^2 + y^2 = 9.

STEP 6

The equation of the circle in standard form is (x+7)2+y2=9(x + 7)^2 + y^2 = 9.

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