find the equation of the circle having center on 3x+y=6 and tangent to both axes
If a circle is tangent to both axes, then its centre lies along the line
y=x (for circle in first or third quadrant) or
y=-x (for circle in second or fourth quadrant).
Make your choice of quadrant(s) and solve the system of equations
3x+y=6
y=x
or
3x+y=6
y=-x
To find the equation of the circle with its center on the line 3x + y = 6 and tangent to both axes, we can follow these steps:
1. Find the equation of the line perpendicular to 3x + y = 6 that passes through the center of the circle. Since the circle is tangent to both axes (x-axis and y-axis), the line passing through its center must be perpendicular to both axes.
The line perpendicular to 3x + y = 6 can be found by evaluating the negative reciprocal of its slope. The slope of the given line is -3 (since it is in the form of y = -3x + 6). So the slope of the perpendicular line is 1/3.
2. Find the coordinates of the center of the circle. We have the equation of the line for the center of the circle, which is perpendicular to the given line. We can solve the two equations simultaneously to find their point of intersection.
Let's solve the system of equations: 3x + y = 6 and y = 1/3x + b, where b is the y-intercept of the perpendicular line.
Substituting y = 1/3x + b into 3x + y = 6, we get:
3x + 1/3x + b = 6
10/3x + b = 6
10x + 3b = 18 (multiplying by 3 to clear the fraction)
From 3x + y = 6, we can solve for y and substitute it into the equation above:
3x + y = 6
y = 6 - 3x
Substituting y = 6 - 3x into 3x + y = 6:
3x + 6 - 3x = 6
6 = 6
The equation 6 = 6 is true, which means that the two equations are dependent (represent the same line) and have infinitely many solutions.
Therefore, the system of equations has infinitely many solutions, indicating that the line 3x + y = 6 passes through the origin (0,0).
So the center of the circle is at the origin (0,0).
3. Compute the radius of the circle. Since the circle is tangent to both axes, the distance from the center (0,0) to any of the axes is equal to the radius of the circle.
The distance from the center (0,0) to the x-axis (y = 0) is the radius. This distance is simply the absolute value of the y-coordinate of the center, which is 0.
Therefore, the radius of the circle is 0.
4. Write the equation of the circle using the center and radius. Since the radius is 0, the equation of the circle will simply be (x - 0)^2 + (y - 0)^2 = 0^2, which simplifies to x^2 + y^2 = 0.
Thus, the equation of the circle with its center on the line 3x + y = 6 and tangent to both axes is x^2 + y^2 = 0.