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Let's solve each of the problems one by one from Exercise 5:

### Problem 1:
**Find the equation of the circle with center at (1, 4) and radius 5.**

The general form of a circle with center \((h, k)\) and radius \(r\) is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]

For this problem:
- Center \((h, k) = (1, 4)\)
- Radius \(r = 5\)

Substituting the values:
\[
(x - 1)^2 + (y - 4)^2 = 5^2
\]
\[
(x - 1)^2 + (y - 4)^2 = 25
\]

### Problem 2:
**Find the equation of the circle with center at \((-1, 7)\) and tangent to the line \(3x - 4y + 6 = 0\).**

A circle tangent to a line has its radius equal to the distance from the center to the line. The formula for the distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is:
\[
d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
\]

For the line \(3x - 4y + 6 = 0\):
- \(A = 3\), \(B = -4\), \(C = 6\)
- Center \((x_1, y_1) = (-1, 7)\)

Substitute into the formula:
\[
d = \frac{|3(-1) - 4(7) + 6|}{\sqrt{3^2 + (-4)^2}} = \frac{| -3 - 28 + 6 |}{\sqrt{9 + 16}} = \frac{|-25|}{5} = \frac{25}{5} = 5
\]

So, the radius \(r = 5\).

Now, the equation of the circle is:
\[
(x + 1)^2 + (y - 7)^2 = 5^2
\]
\[
(x + 1)^2 + (y - 7)^2 = 25
\]

### Problem 3:
**Find the equation of the circle with center \((9, -5)\) and passing through \((3, 4)\).**

To find the radius \(r\), use the distance formula between the center \((9, -5)\) and the point \((3, 4)\):
\[
r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(3 - 9)^2 + (4 - (-5))^2}
\]
\[
r = \sqrt{(-6)^2 + (9)^2} = \sqrt{36 + 81} = \sqrt{117} = 3\sqrt{13}
\]

Now, substitute into the equation:
\[
(x - 9)^2 + (y + 5)^2 = (3\sqrt{13})^2
\]
\[
(x - 9)^2 + (y + 5)^2 = 117
\]

### Problem 4:
**Find the equation of the circle having \((3, 0)\) and \((-2, -4)\) as ends of the diameter.**

The center \((h, k)\) of a circle is the midpoint of the ends of the diameter. Use the midpoint formula:
\[
h = \frac{x_1 + x_2}{2}, \quad k = \frac{y_1 + y_2}{2}
\]
\[
h = \frac{3 + (-2)}{2} = \frac{1}{2}, \quad k = \frac{0 + (-4)}{2} = -2
\]

So, the center is \(\left(\frac{1}{2}, -2\right)\).

To find the radius, use the distance formula between the center and one of the endpoints, say \((3, 0)\):
\[
r = \sqrt{\left(3 - \frac{1}{2}\right)^2 + \left(0 - (-2)\right)^2} = \sqrt{\left(\frac{5}{2}\right)^2 + 2^2}
\]
\[
r = \sqrt{\frac{25}{4} + 4} = \sqrt{\frac{25}{4} + \frac{16}{4}} = \sqrt{\frac{41}{4}} = \frac{\sqrt{41}}{2}
\]

The equation of the circle is:
\[
\left(x - \frac{1}{2}\right)^2 + \left(y + 2\right)^2 = \left(\frac{\sqrt{41}}{2}\right)^2
\]
\[
\left(x - \frac{1}{2}\right)^2 + \left(y + 2\right)^2 = \frac{41}{4}
\]

Or, multiplying through by 4 to remove the fraction:
\[
(2x - 1)^2 + 4(y + 2)^2 = 41
\]

---

### Summary of Solutions:
1. \((x - 1)^2 + (y - 4)^2 = 25\)
2. \((x + 1)^2 + (y - 7)^2 = 25\)
3. \((x - 9)^2 + (y + 5)^2 = 117\)
4. \((2x - 1)^2 + 4(y + 2)^2 = 41\)

Would you like any further details or have any questions about these solutions?

### 5 Related Questions:
1. How do you determine the equation of a circle given three points?
2. What is the geometric interpretation of a circle tangent to a given line?
3. How can you find the intersection points between a circle and a straight line?
4. How is the radius of a circle determined if the circle is described implicitly?
5. How can you convert a circle's general form equation into standard form?

### Tip:
For circles, always remember that the distance from the center to any point on the circumference is the radius, which is useful when the circle is tangent to a given line.

Solve the following problems completely (with figure):
1. Find the equation of the circle with center at (1, 4) and radius 5.
2. Find the equation of the circle with center at (-1, 7) and tangent to the line 3x - 4y + 6 = 0.
3. Find the equation of the circle with center (9, -5) and passing through (3, 4).
4. Find the equation of the circle having (3, 0) and (-2, -4) as ends of the diameter.

Learn to solve circle equations with specified centers, radius, tangents, and diameter endpoints. This exercise covers essential concepts in geometry and analytic geometry, with detailed solutions for finding circle equations given points and lines. Ideal for high school students in Grades 10-12.

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