Find an equation for the circle with center (3,5) tangent to the x-axis .. How?

Answer 1

Since the circle is tangent to the x-axis (the line y = 0), and its centre is at a point having y = 5, its radius will be 5 units. (There is no need to make more heavy weather of deducing this.) Therefore, the equation will be:

(x - 3)^2 + (y - 5)^2 = 5^2 = 25.

This equation basically uses Pythagoras's Theorem to express the following fact : that every point on the circle is at a distance of 5 units from the centre, which is itself at the point (3, 5).

Live long and prosper.

Answer 2

The general equation of a circle is

(x - h)^2 + (y - k)^2 = r^2, where

(h, k) is the center of the circle, and r is the circle's radius.

Since we are given the center of the circle to be (3, 5), plug in h = 3 and k = 5. This will give us

(x - 3)^2 + (y - 5)^2 = r^2

Now, we need to solve for r^2.

In order for us to determine what makes this tangent to the x-axis (which is y = 0), that means, for y = 0, we required exactly one unique solution. Let's let y = 0 and ensure we have one solution.

For the above equation, if y = 0, then we have

(x - 3)^2 + (0 - 5)^2 = r^2
(x - 3)^2 + (-5)^2 = r^2
(x - 3)^2 + 25 = r^2
x^2 - 6x + 9 + 25 = r^2
x^2 - 6x + 34 = r^2
x^2 - 6x + 34 - r^2 = 0

Note that the discriminant of a quadratic determines how many solutions it has; we *want* one solution (by definition of a tangent), and we obtain one solution if the discriminant b^2 - 4ac is equal to 0.

In the above case, a = 1, b = -6, c = (34 - r^2), so
b^2 - 4ac = (-6)^2 - 4(1)(34 - r^2)
= 36 - 4(34 - r^2)
= 36 - 136 + 4r^2
= -100 + 4r^2
= 4r^2 - 100

And we want this equal to 0, so

4r^2 - 100 = 0
r^2 - 25 = 0, so r^2 = 25, giving us r^2, the value we require. Therefore, the equation of our line is

(x - 3)^2 + (y - 5)^2 = 25

Answer 3

The distance from the center to any tangent equals the radius.
In this case the tangent is the x axis ,so the distance from the center is the y coordinate =5 (always taken in absolute value)

so ( x-3)^2+(y-5)^2 = 25

Answer 4

in case you seem on the circle established at (7, -5) you'll see that the tangent element on the x axis should be the point (7,0). the area between (7, -5) and (7,0) is 5 instruments, which signifies that the radius is 5. hence the equation is: (x - 7)^2 + (y + 5)^2 = 5^2 (x - 7)^2 + (y + 5)^2 = 25

Answer 5

The bottom of the circle will intersect with the x axis at (3,0), giving you a circle of radius 5 and center (3,5).