43. What is an equation of the circle that has its center at the origin and is tangent to the line y=3x 7?
To find the equation of a circle that has its center at the origin and is tangent to a given line, you can follow these steps:
1. Determine the slope of the given line: The given line is y = 3x + 7, so the slope is 3.
2. Find the perpendicular slope to the given line: Since the circle is tangent to the line, the line passing through the origin (center of the circle) and perpendicular to the given line will have a slope of the negative reciprocal of the slope of the given line. Therefore, the perpendicular slope is -1/3.
3. Write the equation of the line passing through the origin with the perpendicular slope: Since the circle is tangent to the line y = 3x + 7, it means that it has only one point of contact with the line. This point of contact is the point where the line passing through the origin and perpendicular to the given line intersects the given line. To find this point, we need to solve the system of equations formed by the two lines:
- Line passing through the origin: y = (-1/3)x
- Given line: y = 3x + 7
Solving this system of equations, we find the x-coordinate of the point of contact.
4. Use the distance formula to find the radius of the circle: The distance between the origin (center of the circle) and the point of contact with the line gives us the radius of the circle.
5. Finally, write the equation of the circle using the center and radius: Since the center of the circle is at the origin (0, 0), and the radius is the distance we found in step 4, we can write the equation as (x - h)² + (y - k)² = r², where (h, k) represents the center and r represents the radius.
Following these steps, we can now proceed to find the equation of the circle that satisfies the given conditions.