Find the equation of the circle of radius 
5
5

  tangent to the line 
2
x
+
y
−
3
=
0
2x+y−3=0 and having its center on the line 
2
x
+
3
y
=
4
2x+3y=4.

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 \(2x + 3y = 4\), we can follow these steps:

1. **Determine the center's coordinates**: Let the center of the circle be \((h, k)\). Since the center lies on the line \(2x + 3y = 4\), we can express \(k\) in terms of \(h\):
   \[
   2h + 3k = 4 \quad \Rightarrow \quad k = \frac{4 - 2h}{3}
   \]

2. **Distance from center to the line**: The distance \(d\) from the center \((h, k)\) to the line \(Ax + By + C = 0\) is given by the formula:
   \[
   d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}
   \]
   For the line \(2x + y - 3 = 0\), we have \(A = 2\), \(B = 1\), and \(C = -3\). The distance \(d\) from the center to the line must equal the radius \(r = 5\):
   \[
   d = \frac{|2h + k - 3|}{\sqrt{2^2 + 1^2}} = \frac{|2h + k - 3|}{\sqrt{5}} = 5
   \]
   Therefore, we can set up the equation:
   \[
   |2h + k - 3| = 5\sqrt{5}
   \]

3. **Substituting \(k\)**: Substitute \(k = \frac{4 - 2h}{3}\) into the distance equation:
   \[
   |2h + \frac{4 - 2h}{3} - 3| = 5\sqrt{5}
   \]
   Simplifying inside the absolute value:
   \[
   2h + \frac{4 - 2h}{3} - 3 = \frac{6h + 4 - 2h - 9}{3} = \frac{4h - 5}{3}
   \]
   Thus:
   \[
   \left|\frac{4h - 5}{3}\right| = 5\sqrt{5}
   \]
   Multiplying both sides by 3 gives:
   \[
   |4h - 5| = 15\sqrt{5}
   \]

4. **Solving for \(h\)**: This leads to two cases:
   - \(4h - 5 = 15\sqrt{5}\)
   - \(4h - 5 = -15\sqrt{5}\)

   For the first case:
   \[
   4h = 15\sqrt{5} + 5 \quad \Rightarrow \quad h = \frac{15\sqrt{5} + 5}{4}
   \]
   
   For the second case:
   \[
   4h = -15\sqrt{5} + 5 \quad \Rightarrow \quad h = \frac{-15\sqrt{5} + 5}{4}
   \]

5. **Finding \(k\)**: For each value of \(h\), calculate \(k\):
   - If \(h = \frac{15\sqrt{5} + 5}{4}\):
     \[
     k = \frac{4 - 2\left(\frac{15\sqrt{5} + 5}{4}\right)}{3} = \frac{4 - \frac{30\sqrt{5} + 10}{4}}{3} = \frac{16 - 30\sqrt{5} - 10}{12} = \frac{6 - 30\sqrt{5}}{12} = \frac{1 - 5\sqrt{5}}{2}
     \]

   - If \(h = \frac{-15\sqrt{5} + 5}{4}\):
     \[
     k = \frac{4 - 2\left(\frac{-15\sqrt{5} + 5}{4}\right)}{3} = \frac{16 + 30\sqrt{5} - 10}{12} = \frac{6 + 30\sqrt{5}}{12} = \frac{1 + 5\sqrt{5}}{2}
     \]

6. **Equations of the circles**: The equations of the circles can now be written using \((h, k)\):
   \[
   (x - h)^2 + (y - k)^2 = 25
   \]

   For the first center \(\left(\frac{15\sqrt{5} + 5}{4}, \frac{1 - 5\sqrt{5}}{2}\right)\) and for the second center \(\left(\frac{-15\sqrt{5} + 5}{4}, \frac{1 + 5\sqrt{5}}{2}\right)\).

### Final Equations:
1. First circle:
   \[
   \left(x - \frac{15\sqrt{5} + 5}{4}\right)^2 + \left(y - \frac{1 - 5\sqrt{5}}{2}\right)^2 = 25
   \]

2. Second circle:
   \[
   \left(x - \frac{-15\sqrt{5} + 5}{4}\right)^2 + \left(y - \frac{1 + 5\sqrt{5}}{2}\right)^2 = 25
   \]

If you need further clarifications or have any questions, feel free to ask! Here are some follow-up questions you might consider:

1. Would you like to visualize these circles on a graph?
2. Do you need help with the implications of the circle's tangency to the line?
3. Would you like to explore other properties of circles?
4. Do you want to understand the process of calculating distances to lines better?
5. Are you interested in similar problems involving circles and lines? 

**Tip**: When dealing with geometric problems, sketching the scenario can help clarify relationships and constraints!

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.

Explore how to find the equation of a circle with a radius of 5 that is tangent to the line 2x + y - 3 = 0 and has its center on the line 2x + 3y = 4. This problem integrates concepts of geometry and analytic geometry, suitable for students in Grades 10-12.

Math Problem Statement

tangent to the line 2 x + y − 3

0 2x+y−3=0 and having its center on the line 2 x + 3 y

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