A circle is tangent to the y axis at y=3 and has one x intercept at x=1. find the equation of the circle

To find the equation of the circle, let's break down the given information:

1. **The circle is tangent to the y-axis at \( y = 3 \):**  
   This implies that the center of the circle lies on a horizontal line parallel to the x-axis passing through \( y = 3 \). Since the circle is tangent to the y-axis, the distance from the center to the y-axis is equal to the radius \( r \), meaning the x-coordinate of the center is the radius \( r \).

2. **The circle has one x-intercept at \( x = 1 \):**  
   This means that the circle passes through the point \( (1, 0) \).

Let the center of the circle be \( (r, 3) \), where \( r \) is the radius of the circle.

The general equation of a circle is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
where \( (h, k) \) is the center and \( r \) is the radius.

In our case, the center is \( (r, 3) \), so the equation becomes:
\[
(x - r)^2 + (y - 3)^2 = r^2
\]

Now, substitute the point \( (1, 0) \) into this equation, since the circle passes through this point:
\[
(1 - r)^2 + (0 - 3)^2 = r^2
\]
\[
(1 - r)^2 + 9 = r^2
\]
Expand and solve for \( r \):
\[
(1 - r)^2 = 1 - 2r + r^2
\]
Substituting this back into the equation:
\[
1 - 2r + r^2 + 9 = r^2
\]
Simplify:
\[
1 - 2r + 9 = 0
\]
\[
10 - 2r = 0
\]
\[
r = 5
\]

Thus, the radius \( r = 5 \), and the center of the circle is \( (5, 3) \).

The equation of the circle is:
\[
(x - 5)^2 + (y - 3)^2 = 25
\]

### Final Answer:
The equation of the circle is:
\[
(x - 5)^2 + (y - 3)^2 = 25
\]

Would you like further details on any step, or do you have any questions?

Here are 5 related questions:
1. How do you find the equation of a circle given the center and a point on the circle?
2. What is the geometric significance of tangency between a circle and a line?
3. How can you determine the radius of a circle if you know its equation?
4. What is the effect of shifting a circle horizontally or vertically on its equation?
5. How can you solve for the center and radius of a circle from its general equation?

**Tip:** When solving circle-related problems, always remember that the distance from the center to any point on the circle is the radius. This fact helps in setting up equations based on the circle's properties.

A circle is tangent to the y axis at y=3 and has one x intercept at x=1. Find the equation of the circle.

Discover how to find the equation of a circle tangent to the y-axis at y=3 and with an x-intercept at x=1. This solution covers essential geometric concepts, such as the equation of a circle, tangent properties, and solving for the radius. Suitable for students in Grades 9-11.

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