find the equation of the circle of radius √5 tangent to the line 2x+y-3=0 and having its center on the line 2x+3y=4

To find the equation of the circle with radius \( \sqrt{5} \), tangent to the line \( 2x + y - 3 = 0 \) and having its center on the line \( 2x + 3y = 4 \), we can follow these steps:

1. **Identify the line equations**:
   - The first line (the tangent line) is \( 2x + y - 3 = 0 \).
   - The second line (on which the center lies) is \( 2x + 3y - 4 = 0 \).

2. **Find the distance from the center of the circle to the tangent line**:
   The distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
   \[
   d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}
   \]
   For the line \( 2x + y - 3 = 0 \), \( A = 2 \), \( B = 1 \), and \( C = -3 \). The radius of the circle is \( \sqrt{5} \).

   Therefore, we have:
   \[
   \frac{|2x_0 + y_0 - 3|}{\sqrt{2^2 + 1^2}} = \sqrt{5}
   \]
   Simplifying, we find:
   \[
   |2x_0 + y_0 - 3| = \sqrt{5} \cdot \sqrt{5} = 5
   \]
   Thus:
   \[
   2x_0 + y_0 - 3 = 5 \quad \text{or} \quad 2x_0 + y_0 - 3 = -5
   \]
   This leads to two cases:
   - Case 1: \( 2x_0 + y_0 = 8 \)
   - Case 2: \( 2x_0 + y_0 = -2 \)

3. **Substitute into the second line equation**:
   The center \( (x_0, y_0) \) must also satisfy \( 2x_0 + 3y_0 = 4 \).

   **For Case 1:**
   - From \( 2x_0 + y_0 = 8 \), we can express \( y_0 \) as \( y_0 = 8 - 2x_0 \).
   - Substituting into the second line equation:
     \[
     2x_0 + 3(8 - 2x_0) = 4
     \]
     \[
     2x_0 + 24 - 6x_0 = 4 \implies -4x_0 + 24 = 4 \implies -4x_0 = -20 \implies x_0 = 5
     \]
   - Then:
     \[
     y_0 = 8 - 2(5) = -2
     \]
   - So, the center is \( (5, -2) \).

   **For Case 2:**
   - From \( 2x_0 + y_0 = -2 \), we have \( y_0 = -2 - 2x_0 \).
   - Substituting into the second line equation:
     \[
     2x_0 + 3(-2 - 2x_0) = 4
     \]
     \[
     2x_0 - 6 - 6x_0 = 4 \implies -4x_0 - 6 = 4 \implies -4x_0 = 10 \implies x_0 = -\frac{5}{2}
     \]
   - Then:
     \[
     y_0 = -2 - 2\left(-\frac{5}{2}\right) = -2 + 5 = 3
     \]
   - So, the center is \( \left(-\frac{5}{2}, 3\right) \).

4. **Find the equations of the circles**:
   The equation of a circle with center \( (h, k) \) and radius \( r \) is given by:
   \[
   (x - h)^2 + (y - k)^2 = r^2
   \]
   - **For center \( (5, -2) \)**:
     \[
     (x - 5)^2 + (y + 2)^2 = 5
     \]
   - **For center \( \left(-\frac{5}{2}, 3\right) \)**:
     \[
     \left(x + \frac{5}{2}\right)^2 + (y - 3)^2 = 5
     \]

Thus, the two possible equations of the circles are:
1. \( (x - 5)^2 + (y + 2)^2 = 5 \)
2. \( \left(x + \frac{5}{2}\right)^2 + (y - 3)^2 = 5 \)

Would you like any further details or have questions about this problem? Here are some related questions for further exploration:

1. How can we find the point of tangency on the line?
2. What is the significance of the center being on a specific line?
3. How does the radius affect the position of the circle relative to the tangent line?
4. Can you derive the slope of the tangent line?
5. How do you determine if two circles intersect or are tangent? 

**Tip**: Always check the conditions of tangency and distance when dealing with circles and lines!

Explore how to find the equation of a circle with radius √5 that is tangent to the line 2x + y - 3 = 0 and has its center on the line 2x + 3y = 4. This detailed solution walks through the necessary steps and formulas for solving geometry problems involving circles and tangents. Suitable for students in Grades 10-12.

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