# a circle is tangent to the y-axis at y=3 and has one x-intercept at x=1

a) what's the other x-intercept
b) what's the equation

i have no clue how to solve this!

You know that the center of the circle is somewhere in the line y = 3, denote the x-coordinate by rx. Then the radius of the circle is |rx|. The distance from the center to the point x = 1 on the x-axis must be equal to the radius, which is |rx|. Therefore:

(rx-1)^2 + 9 = rx^2 --->

rx = 5.

Equation of the circle is thus:

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

And you see that the other x intercept is at x = 9

## To solve this problem, follow these steps:

a) Determine the x-coordinate of the center of the circle:

Since the circle is tangent to the y-axis at y=3, the y-coordinate of the center is 3. Let's denote the x-coordinate of the center as cx.

b) Use the fact that the center of the circle is equidistant from both x-intercepts:

The distance from the center (cx,3) to the x-intercept (1,0) is equal to the distance from the center to the other x-intercept (let's call it dx, 0).

Using the distance formula, the equation becomes:

sqrt((1-cx)^2 + (0-3)^2) = sqrt((dx-cx)^2 + (0-3)^2)

Expanding and simplifying:

sqrt((1-cx)^2 + 9) = sqrt((dx-cx)^2 + 9)

Squaring both sides to eliminate the square root:

(1-cx)^2 + 9 = (dx-cx)^2 + 9

Simplifying further:

1 - 2cx + cx^2 + 9 = dx^2 - 2dxcx + cx^2 + 9

Canceling out 9 on both sides:

1 - 2cx + cx^2 = dx^2 - 2dxcx + cx^2

Simplifying further:

1 - 2cx = dx^2 - 2dxcx

Factoring out cx:

1 - dx^2 = (2cx - dx)(-2cx)

Since (2cx - dx) and (-2cx) are equal and opposite, we can divide both sides by -2cx:

(dx^2 - 1) / (2cx - dx) = 1

Solving for dx:

dx^2 - 1 = 2cx - dx

dx^2 + dx - 2cx + 1 = 0

Now we can solve this quadratic equation for dx using factoring or the quadratic formula.

## To solve this problem, we can use the concept of tangents and the equation of a circle.

a) To find the other x-intercept, we need to find the x-coordinate of the center of the circle. Since the circle is tangent to the y-axis at y = 3, the center lies on the line y = 3. Let's denote the x-coordinate of the center as rx.

We know that the distance from the center to the point x = 1 on the x-axis is equal to the radius of the circle. Since the circle is tangent to the y-axis, the distance from the center to the point (1, 0) is the radius. Using the distance formula, we have:

√[ (rx - 1)^2 + (3 - 0)^2 ] = |rx|

Simplifying this equation, we get:

(rx - 1)^2 + 9 = rx^2

Expanding and canceling the rx^2 terms, we have:

rx^2 - 2rx + 1 + 9 = rx^2

Simplifying further, we get:

-2rx + 10 = 0

Adding 2rx to both sides and dividing by 2, we find:

rx = 5

Therefore, the x-coordinate of the center of the circle is 5. Knowing that the circle passes through (1, 0), we can determine the other x-intercept by substituting y = 0 into the equation of the circle:

(x - 5)^2 + (0 - 3)^2 = 25

Simplifying, we have:

(x - 5)^2 + 9 = 25

(x - 5)^2 = 16

Taking the square root of both sides, we get:

x - 5 = ±4

Solving for x, we have two solutions:

x - 5 = 4 --> x = 9
x - 5 = -4 --> x = 1

Therefore, the other x-intercept is x = 9.

b) To find the equation of the circle, we have already determined the x-coordinate of the center as 5. Using the formula of a circle with center (h, k) and radius r, which is given by:

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

We substitute the values of the center (5, 3) and the radius (5), and we get:

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

Hence, the equation of the circle is (x - 5)^2 + (y - 3)^2 = 25.