Find the equation of the circle determined by the given conditions
1.The circle is tangent to the x- and y-axes and passes through the point (6, 3).

To find the equation of the circle that is tangent to both the x-axis and y-axis and passes through the point \((6, 3)\), we follow these steps:

### Step 1: Identify the center of the circle
Since the circle is tangent to both the x-axis and y-axis, the center of the circle must be equidistant from both axes. This implies that the center \((h, k)\) of the circle has equal x and y coordinates, i.e., \(h = k\).

Let the center of the circle be \((h, h)\).

### Step 2: Determine the radius
The radius \(r\) of the circle is the distance from the center to either axis, which is \(h\) since \(h\) is the distance from the center to the x-axis and y-axis.

### Step 3: Use the given point
The point \((6, 3)\) lies on the circle, so the distance from this point to the center \((h, h)\) should be equal to the radius \(h\). The distance formula between the point \((6, 3)\) and the center \((h, h)\) is:

\[
\sqrt{(6 - h)^2 + (3 - h)^2} = h
\]

### Step 4: Solve for \(h\)
Square both sides of the equation to remove the square root:

\[
(6 - h)^2 + (3 - h)^2 = h^2
\]

Expanding the squares:

\[
(36 - 12h + h^2) + (9 - 6h + h^2) = h^2
\]

Combine like terms:

\[
45 - 18h + 2h^2 = h^2
\]

Simplify the equation:

\[
h^2 - 18h + 45 = 0
\]

### Step 5: Solve the quadratic equation
Use the quadratic formula \(h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -18\), and \(c = 45\):

\[
h = \frac{18 \pm \sqrt{(-18)^2 - 4(1)(45)}}{2(1)}
\]

\[
h = \frac{18 \pm \sqrt{324 - 180}}{2}
\]

\[
h = \frac{18 \pm \sqrt{144}}{2}
\]

\[
h = \frac{18 \pm 12}{2}
\]

This gives two possible values for \(h\):

\[
h = \frac{18 + 12}{2} = 15 \quad \text{or} \quad h = \frac{18 - 12}{2} = 3
\]

### Step 6: Determine the correct value of \(h\)
Since the radius \(h\) must be the distance from the center to the axes and must satisfy the point \((6, 3)\), \(h = 3\) is the correct value (as \(h = 15\) would not place the center near enough to satisfy the equation with the point \((6, 3)\)).

### Step 7: Write the equation of the circle
The center is \((3, 3)\) and the radius \(r = 3\), so the equation of the circle is:

\[
(x - 3)^2 + (y - 3)^2 = 3^2
\]

\[
(x - 3)^2 + (y - 3)^2 = 9
\]

This is the equation of the circle.

Would you like further details or have any questions? Here are some related problems you might find interesting:

1. Find the equation of a circle that is tangent to both the x-axis and the y-axis and passes through the point \((-4, 2)\).
2. Determine the equation of the circle with center \((2, 5)\) and radius 7.
3. What is the equation of a circle tangent to the line \(y = 2\) with center at \((4, 6)\)?
4. Find the equation of the circle that passes through \((1, 1)\), \((4, 5)\), and \((7, 3)\).
5. Determine the equation of a circle that is tangent to the y-axis at the point \((0, 5)\) and has a radius of 7.

**Tip:** When dealing with circles tangent to axes, remember that the distance from the center to the axis equals the radius.

Learn how to find the equation of a circle that is tangent to both the x-axis and y-axis, and passes through the point (6, 3). This problem involves geometry concepts, including circle equations and the distance formula. Suitable for high school students.

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