find the equation of the circles described. tangent to 2x + 3y + 13 = 0 and 2x - 3y - 1 = 0 and contains (0,4)

Let L1 be the line 2x + 3y + 13 = 0. Its slope is -2/3

Let L2 be the line 2x - 3y - 1 = 0. It has slope 2/3.

So, the bisectors of the angles formed by L1,L2 are horizontal and vertical. Since (0,4) is above both lines, we want the vertical bisector x = -3.

So, we want a circle whose center is (-3,k) which passes through (0,4) and is tangent to L1 (and hence L2).

The distance from (-3,k) to L1 is (3k+7)/√13

So far we have the circle

(x+3)^2 + (y-k)^2 = (3k+7)^2/13

We know the circle passes through (0,4), so

3^2 + (4-k)^2 = (3k+7)^2/13
k=2

Looks like the circle is
(x+3)^2 + (y-2)^2 = 13

To find the equation of a circle, we need its center coordinates (h, k) and its radius r.

Step 1: Find the intersection point of the two given lines.
To find the intersection point, we need to solve the system of equations formed by the two lines:

2x + 3y + 13 = 0 ...(1)
2x - 3y - 1 = 0 ...(2)

Multiplying equation (1) by 3 and equation (2) by 3, we get:
6x + 9y + 39 = 0 ...(3)
6x - 9y - 3 = 0 ...(4)

Now, adding equations (3) and (4), we eliminate the term "x" and solve for "y":
18y + 36 = 0
18y = -36
y = -2

Substituting the value of y in equation (1), we find the corresponding x-coordinate:
2x + 3(-2) + 13 = 0
2x - 6 + 13 = 0
2x + 7 = 0
2x = -7
x = -7/2

So, the intersection point of the two given lines is (-7/2, -2).

Step 2: Find the distance from the center to the given point (0, 4).
To find the radius, we need to find the distance between the center of the circle and the given point.

Using the distance formula:
Distance between two points (x1, y1) and (x2, y2) is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Using (h, k) as the center coordinates and (0, 4) as the given point, the distance is:
d = sqrt((0 - h)^2 + (4 - k)^2)

Step 3: Write the equation using the center coordinates and the radius.
The equation of a circle with center coordinates (h, k) and radius r is given by:
(x - h)^2 + (y - k)^2 = r^2

Substituting the values in the equation:
(-7/2 - h)^2 + (-2 - k)^2 = d^2

So, the equation of the circle tangent to the lines 2x + 3y + 13 = 0 and 2x - 3y - 1 = 0 and contains the point (0, 4) is given by:
((-7/2 - h)^2 + (-2 - k)^2) = (sqrt((0 - h)^2 + (4 - k)^2))^2

To find the equation of a circle, we need to know its center coordinates and its radius.

In this case, we are given that the circle is tangent to the lines 2x + 3y + 13 = 0 and 2x - 3y - 1 = 0. Let's find the point of tangency on each line.

For the first line, 2x + 3y + 13 = 0, we can rearrange it to the slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept. Thus, we have 3y = -2x - 13, or y = (-2/3)x - 13/3.

Since the circle is tangent to this line, the radius will be perpendicular to the line at the point of tangency. The slope of the tangent line is the negative reciprocal of the slope of the given line, which is 3/2.

Using the point-slope form of a line, we can find the equation of a line passing through (0, 4) with a slope of 3/2.

y - 4 = (3/2)x

We can rearrange the equation to slope-intercept form:

y = (3/2)x + 4

Now, let's find the point of tangency on the second line, 2x - 3y - 1 = 0.

Rearranging the equation gives us 3y = 2x + 1, or y = (2/3)x + 1/3.

Similarly, since the circle is tangent to this line, the radius will be perpendicular to the line at the point of tangency. The slope of the tangent line is the negative reciprocal of the slope of the given line, which is -3/2.

Using the point-slope form of a line, we can find the equation of a line passing through (0, 4) with a slope of -3/2.

y - 4 = (-3/2)x

Again, we can rearrange the equation to slope-intercept form:

y = (-3/2)x + 4

Next, we need to find the point of intersection of these two lines. We can solve the equations of the lines to find their intersection point.

Setting the two equations equal to each other:

(3/2)x + 4 = (-3/2)x + 4

Simplifying the equation:

3x + 12 = -3x + 8

Combining like terms:

6x = -4

Dividing both sides of the equation by 6:

x = -4/6 = -2/3

Substituting the value of x into one of the equations:

y = (2/3)(-2/3) + 1/3

Simplifying the equation:

y = -4/9 + 1/3 = -4/9 + 3/9 = -1/9

Therefore, the point of intersection is (-2/3, -1/9), which represents the center of the circle.

Now, let's find the radius of the circle. We can use the distance formula to find the distance between the center of the circle and the given point (0, 4).

d = √[(x2 - x1)² + (y2 - y1)²]

Substituting the values:

d = √[((-2/3) - 0)² + ((-1/9) - 4)²]

Simplifying the equation:

d = √[(-2/3)² + (-1/9 - 36/9)²] = √[(4/9) + (-37/9)²] = √[41/9]

Therefore, the radius of the circle is √(41/9).

Now we can write the equation of the circle using the center coordinates and the radius.

The equation of the circle is:

(x - (-2/3))² + (y - (-1/9))² = (√(41/9))²

(x + 2/3)² + (y + 1/9)² = 41/9

After simplification, we can write the final equation:

9(x + 2/3)² + 9(y + 1/9)² = 41