finding equation to a circle

Question

Find an equation of the circle that satisfies the given conditions. (Use the variables x and y.)

Center (7, −1);    tangent to the x-axis

Answer 1

The formula for a circle with centered at (a,b) with a radius of r is:

r = √((x-a)^2+(y-b)^2)

It's given that the center is (7, -1), so we can plug in a = 7 and b = -1:

r = √((x-7)^2+(y+1)^2)

Now you only need to find the radius, (the distance from the center of the circle to every point on the circle).

A tangent line on a circle is a line that intersects the circle at only one point, called the tangent point and runs perpendicular to the line from the center of the circle to the tangent point.

Since the x-axis (horizontal) is a tangent line, the line running from the center (7, -1) to the tangent point is vertical, meaning they have the same x-value. So for the tangent point x=7, and since the tangent line is on the x-axis we know that y=0 (the x-axis is all points satisfying y=0). So the point (7,0) is a point on our circle.

Using the distance formula, we have that the distance from (7, -1) to (7,0), our radius, is

r = √((7-7)^2+(-1-0)^2)

r = √1

r = 1

So our equation is

1 = √((x-7)^2+(y+1)^2)

You could also have plugged the point (7,0) with x=7 y=0 to

r = √((x-7)^2+(y+1)^2)

but thinking about this geometrically in this case is a little faster.

Answer 2

The equation is. (X-7)^2 + (y+1)^2 = 1

By inspection, you know the radius is. 1 since -1 is 1 unit from the x axis which is the tangent reference!