Prove that the circle `x^(2) +y^(2) - 6 x -2 y + 9 = 0`
(i) touches the x-axis, (ii) lies entirely inside the circle `x^(2) + y^(2) = 18`.

Answer

To solve the problem, we need to prove two things about the circle given by the equation \(x^2 + y^2 - 6x - 2y + 9 = 0\):

1. It touches the x-axis.
2. It lies entirely inside the circle given by the equation \(x^2 + y^2 = 18\).

### Step 1: Rewrite the Circle's Equation

We start with the equation of the circle:

\[ x^2 + y^2 - 6x - 2y + 9 = 0 \]

We can rearrange this equation to identify the center and the radius of the circle.

### Step 2: Complete the Square

To find the center and radius, we complete the square for the \(x\) and \(y\) terms.

For \(x\):
\[ x^2 - 6x \quad \text{can be written as} \quad (x - 3)^2 - 9 \]

For \(y\):
\[ y^2 - 2y \quad \text{can be written as} \quad (y - 1)^2 - 1 \]

Now substituting these back into the equation:

\[ (x - 3)^2 - 9 + (y - 1)^2 - 1 + 9 = 0 \]

This simplifies to:

\[ (x - 3)^2 + (y - 1)^2 - 1 = 0 \]

Thus, we have:

\[ (x - 3)^2 + (y - 1)^2 = 1 \]

### Step 3: Identify the Center and Radius

From the equation \((x - 3)^2 + (y - 1)^2 = 1\), we can identify:

- Center: \(C(3, 1)\)
- Radius: \(r = 1\)

### Step 4: Prove that the Circle Touches the X-axis

The circle touches the x-axis if the distance from the center to the x-axis is equal to the radius.

The distance from the center \(C(3, 1)\) to the x-axis (which is \(y = 0\)) is simply the \(y\)-coordinate of the center:

\[ \text{Distance} = 1 \]

Since the radius of the circle is also \(1\), we conclude that the circle touches the x-axis.

### Conclusion for Part (i)

Thus, we have proved that the circle touches the x-axis.

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### Step 5: Prove that the Circle Lies Entirely Inside the Circle \(x^2 + y^2 = 18\)

Now, we need to check if the circle lies entirely inside the circle given by:

\[ x^2 + y^2 = 18 \]

### Step 6: Identify the Center and Radius of the Second Circle

The equation \(x^2 + y^2 = 18\) can be rewritten as:

\[ x^2 + y^2 = (3\sqrt{2})^2 \]

Thus, the center of this circle is at \(O(0, 0)\) and the radius is \(R = 3\sqrt{2}\).

### Step 7: Calculate the Distance from the Origin to the Center of the First Circle

The distance from the origin \(O(0, 0)\) to the center of the first circle \(C(3, 1)\) is:

\[ d = \sqrt{(3 - 0)^2 + (1 - 0)^2} = \sqrt{9 + 1} = \sqrt{10} \]

### Step 8: Check if the First Circle Lies Inside the Second Circle

For the first circle to lie entirely inside the second circle, the following condition must hold:

\[ d + r < R \]

Substituting the values we found:

\[ \sqrt{10} + 1 < 3\sqrt{2} \]

### Step 9: Calculate \(3\sqrt{2}\)

Calculating \(3\sqrt{2}\):

\[ 3\sqrt{2} \approx 3 \times 1.414 = 4.242 \]

### Step 10: Compare Values

Now we need to check if:

\[ \sqrt{10} + 1 < 4.242 \]

Calculating \(\sqrt{10} \approx 3.162\):

\[ 3.162 + 1 = 4.162 < 4.242 \]

### Conclusion for Part (ii)

Thus, we have proved that the first circle lies entirely inside the second circle.

### Final Conclusion

1. The circle touches the x-axis.
2. The circle lies entirely inside the circle \(x^2 + y^2 = 18\).

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