Find an equation(s) of the circle(s) described. Tangent to x + y = 0 and x - y - 6 = 0; center on 3x - y + 3 = 0. How do you solve this?

To find the equation(s) of the circle(s) described, we can start by determining the center of the circle.

Given that the center of the circle lies on the line 3x - y + 3 = 0, we can find the coordinates of the center by solving this equation for x and y.

Step 1: Solve 3x - y + 3 = 0 for y:
y = 3x + 3

Step 2: Substitute this value for y into the equation of the line x + y = 0:
x + (3x + 3) = 0
4x + 3 = 0
4x = -3
x = -3/4

Step 3: Substitute the value of x back into the equation y = 3x + 3 to find y:
y = 3(-3/4) + 3
y = -9/4 + 12/4
y = 3/4

Therefore, the center of the circle is (x, y) = (-3/4, 3/4).

Next, we need to determine the radius of the circle. The radius will be the perpendicular distance from the center of the circle to one of the given lines, such as x + y = 0.

Step 4: Finding the perpendicular distance between the center of the circle and the line x + y = 0:
To find the perpendicular distance between a point (x1, y1) and a line Ax + By + C = 0, we can use the formula:

Distance = |Ax1 + By1 + C| / sqrt(A^2 + B^2)

Using the point (-3/4, 3/4) and the line x + y = 0, we have:

Distance = |(1)(-3/4) + (1)(3/4) + 0| / sqrt(1^2 + 1^2)
Distance = |-3/4 + 3/4| / sqrt(2)
Distance = 0 / sqrt(2)
Distance = 0

Since the distance is 0, the circle is tangent to the line x + y = 0. Similarly, we can find the distance between the center and the line x - y - 6 = 0, which should also be 0 since the circle is tangent to that line as well.

As a result, there is only one circle that satisfies the given conditions, and the equation of the circle with the center at (-3/4, 3/4) and radius 0 is simply:

(x + 3/4)^2 + (y - 3/4)^2 = 0