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In the given figure, we need to find the equations for both the circle and the line.

### 1. **Equation of the Circle:**
The circle is centered at the origin, \( (0, 0) \), and passes through the point \( (-5, 12) \).

The general equation of a circle with center \( (h, k) \) and radius \( r \) is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Since the center is at \( (0, 0) \), the equation simplifies to:
\[
x^2 + y^2 = r^2
\]

We can determine \( r \) by finding the distance from the center \( (0, 0) \) to the point \( (-5, 12) \), which lies on the circle:
\[
r = \sqrt{(-5 - 0)^2 + (12 - 0)^2} = \sqrt{25 + 144} = \sqrt{169} = 13
\]

Thus, the equation of the circle is:
\[
x^2 + y^2 = 13^2 = 169
\]
\[
x^2 + y^2 = 169
\]

### 2. **Equation of the Line:**
The line is passing through the point \( (-5, 12) \) and is perpendicular to the radius of the circle at that point. The slope of the radius from the center \( (0, 0) \) to \( (-5, 12) \) is:
\[
\text{slope of radius} = \frac{12 - 0}{-5 - 0} = -\frac{12}{5}
\]

Since the line is perpendicular to the radius, its slope will be the negative reciprocal of \( -\frac{12}{5} \), which is:
\[
\text{slope of line} = \frac{5}{12}
\]

Now, using the point-slope form of a line equation:
\[
y - y_1 = m(x - x_1)
\]
Substituting \( (-5, 12) \) and the slope \( \frac{5}{12} \):
\[
y - 12 = \frac{5}{12}(x + 5)
\]

Expanding this equation:
\[
y - 12 = \frac{5}{12}x + \frac{25}{12}
\]
Multiplying the entire equation by 12 to eliminate the fraction:
\[
12y - 144 = 5x + 25
\]
Rearranging:
\[
5x - 12y = -169
\]

Thus, the equation of the line is:
\[
5x - 12y = -169
\]

### Summary:
- The equation of the **circle** is: \( x^2 + y^2 = 169 \)
- The equation of the **line** is: \( 5x - 12y = -169 \)

Would you like more details or have any other questions?

Here are some related questions:
1. How do you determine the radius of a circle given its center and a point on the circle?
2. What is the significance of the negative reciprocal in the slope of perpendicular lines?
3. How do you derive the general equation of a line from the point-slope form?
4. What other geometric properties can be derived from a circle and a tangent line?
5. How can you verify if a point lies on both the circle and the line?

**Tip**: Always check if a line is tangent to a circle by verifying that the distance between the center of the circle and the line equals the radius of the circle.

Find equations for the circle and the line in the given figure, where the circle is centered at the origin and passes through the point (-5, 12), and the line is perpendicular to the radius at that point.

Learn how to find the equations of a circle centered at the origin and passing through a given point, as well as the equation of a line perpendicular to the radius. This problem covers key concepts in coordinate geometry, including the circle equation and perpendicular slopes. Suitable for students in Grades 9-11.

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