How would the equations change if the point (8,2)(8,2) did not lie on both circles

If the point \((8, 2)\) did not lie on both circles, we need to follow a different approach to find the equations of the two circles. We still use the given conditions about the common tangent and the centers on the line \(x + 2y = 22\), but without the constraint that the point \((8, 2)\) lies on both circles. Here’s how we can proceed:

### Step 1: Determine the Centers of the Circles

1. **Line of centers**: The centers of the circles lie on the line \(x + 2y = 22\).
2. **Common tangent**: The x-axis is a common tangent to the circles. Therefore, the y-coordinate of the centers is equal to the radii of the circles.

Let the centers be \((h_1, k_1)\) and \((h_2, k_2)\) where \(k_1\) and \(k_2\) are the radii of the circles.

### Step 2: Use the Line Equation for Centers

From the line equation:
- For the first center: \(h_1 + 2k_1 = 22\)
- For the second center: \(h_2 + 2k_2 = 22\)

Therefore:
\[ h_1 = 22 - 2k_1 \]
\[ h_2 = 22 - 2k_2 \]

### Step 3: Write the Equations of the Circles

The general equation of a circle is:
\[ (x - h)^2 + (y - k)^2 = k^2 \]

So, the equations for the two circles become:
\[ (x - (22 - 2k_1))^2 + (y - k_1)^2 = k_1^2 \]
\[ (x - (22 - 2k_2))^2 + (y - k_2)^2 = k_2^2 \]

### Step 4: Conditions for the Tangent Line \(4x + 3y = 88\)

To prove that the line \(4x + 3y = 88\) is a common tangent, we need to ensure that the distance from each center to this line equals the respective radius.

#### Distance from a Point to a Line Formula

The distance \(d\) from a point \((h, k)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}} \]

For the line \(4x + 3y = 88\), we rewrite it as \(4x + 3y - 88 = 0\).

### Step 5: Distance Calculation

For the first circle with center \((22 - 2k_1, k_1)\):
\[ d_1 = \frac{|4(22 - 2k_1) + 3k_1 - 88|}{\sqrt{4^2 + 3^2}} = \frac{|88 - 8k_1 + 3k_1 - 88|}{5} = \frac{|-5k_1|}{5} = k_1 \]

For the second circle with center \((22 - 2k_2, k_2)\):
\[ d_2 = \frac{|4(22 - 2k_2) + 3k_2 - 88|}{\sqrt{4^2 + 3^2}} = \frac{|88 - 8k_2 + 3k_2 - 88|}{5} = \frac{|-5k_2|}{5} = k_2 \]

Since the distance equals the radius for both circles, the line \(4x + 3y = 88\) is indeed a common tangent to both circles.

### Final Equations of the Circles

The final equations of the two circles are:
\[ (x - (22 - 2k_1))^2 + (y - k_1)^2 = k_1^2 \]
\[ (x - (22 - 2k_2))^2 + (y - k_2)^2 = k_2^2 \]

where \(k_1\) and \(k_2\) are the radii of the respective circles, and they can be any positive values that satisfy the conditions.

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### Questions
1. What are the properties of a circle with a given tangent line?
2. How do you find the radius of a circle given its center and a tangent line?
3. What role does the line \(x + 2y = 22\) play in determining the circle's equations?
4. How would the problem change if the y-axis was the common tangent instead of the x-axis?
5. What is the significance of the point \((8, 2)\) in the original problem?
6. Can you derive the equation of a circle given two tangents?
7. How can you determine if a line is tangent to a given circle?
8. What other methods can be used to find the equation of a circle with specific properties?

### Tip
When dealing with circle problems, always remember to verify that the given conditions hold for your derived equations. This helps to ensure the accuracy of your solutions.

Explore how to determine the equations of circles and verify tangent lines using advanced geometry concepts. Learn how to solve problems involving circle geometry with detailed explanations and examples. Suitable for advanced high school students and beyond.

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