Generate an illustrative image showing a 2D Cartesian plane. On the plane, show a circle that passes through the point (-4,-4). Also, illustrate a straight line represented by the equation 2x-3y+9=0. Make sure the circle is perfectly tangent to the line at the point (-3,1). Ensure everything except the numbers and symbols are colored and no text is included in your art.

What is the equation of a circle passing through (-4,-4) and tangent to the line 2x-3y+9=0 at (-3,1)?

how do you get the center?

Why did the circle become friends with the line? Because they had a tangent-ial connection!

But in all seriousness, let's find the equation of the circle.
To start, we need to find the center of the circle. Since it is tangent to the line at (-3, 1), the distance between the center and the line is the radius of the circle.

The distance between a point and a line can be found using the formula:
distance = |Ax + By + C| / √(A^2 + B^2)

In this case, the line equation is: 2x - 3y + 9 = 0
So we have A = 2, B = -3, and C = 9. Substituting these values into the formula, we get:
distance = |2(-3) - 3(1) + 9| / √(2^2 + (-3)^2)
= |(-6) - 3 + 9| / √(4 + 9)
= |0| / √(13)
= 0 / √13
= 0

Since the distance between the center and the line is 0, the line passes through the center of the circle. Thus, the center of the circle is (-3, 1).

Next, we can use the center (-3, 1) and one of the given points on the circle (-4, -4) to find the radius of the circle.

The distance between two points can be found using the formula:
distance = √[(x2 - x1)^2 + (y2 - y1)^2]

Substituting the points (-3, 1) and (-4, -4) into the formula, we get:
radius = √[(-4 - (-3))^2 + (-4 - 1)^2]
= √[(-4 + 3)^2 + (-4 - 1)^2]
= √[(-1)^2 + (-5)^2]
= √[1 + 25]
= √26

Finally, with the center (-3, 1) and the radius √26, we can write the equation of the circle in standard form:

(x - h)^2 + (y - k)^2 = r^2

Substituting the values, we have:
(x - (-3))^2 + (y - 1)^2 = (√26)^2
(x + 3)^2 + (y - 1)^2 = 26

So, the equation of the circle passing through (-4, -4) and tangent to the line 2x - 3y + 9 = 0 at (-3, 1) is:
(x + 3)^2 + (y - 1)^2 = 26

To find the equation of a circle passing through (-4,-4) and tangent to the line 2x-3y+9=0 at (-3,1), we need to first find the center and radius of the circle.

Step 1: Find the center of the circle
Since the circle passes through (-4,-4), the center of the circle will be equidistant from (-4,-4) and the point where the circle is tangent to the line.

Let's find the coordinates of the point of tangency. First, let's rewrite the equation of the line in slope-intercept form: 2x - 3y + 9 = 0
-3y = -2x - 9
y = (2/3)x + 3

The slope of this line is 2/3. Since the circle is tangent to the line at (-3,1), then the line perpendicular to the tangent will have a slope of -3/2 (the negative reciprocal of 2/3).

Using the point-slope form of a line, we can find the equation of the perpendicular line passing through (-3,1):
y - 1 = (-3/2)(x - (-3))
y - 1 = (-3/2)(x + 3)
y - 1 = (-3/2)x - 9/2
y = (-3/2)x - 7/2

Now, we solve the system of equations:
(-4, -4) is equidistant from (-3,1) and the center of the circle. So, we can set up the equation:

√[(x - (-4))^2 + (y - (-4))^2] = √[(x - (-3))^2 + (y - 1)^2]

Solving this equation, we get:
√[(x + 4)^2 + (y + 4)^2] = √[(x + 3)^2 + (y - 1)^2]

Squaring both sides, we have:
(x + 4)^2 + (y + 4)^2 = (x + 3)^2 + (y - 1)^2

Expanding and simplifying this equation,
x^2 + 8x + 16 + y^2 + 8y + 16 = x^2 + 6x + 9 + y^2 - 2y + 1

Rearranging the terms,
8x + 8y + 32 = 6x - 2y + 10

Simplifying,
2x + 10y + 22 = 0

Step 2: Find the radius of the circle
The radius of the circle is the distance between the center and any point on the circle. We can use the distance formula to find the distance between the center (-h,-k) and (-4,-4):

r = √[(x2 - x1)^2 + (y2 - y1)^2]
r = √[(-4 - (-4))^2 + (-4 - (-4))^2]
r = √[(0)^2 + (0)^2]
r = √[0 + 0]
r = √0
r = 0

Therefore, the radius of the circle is 0.

Step 3: Write the equation of the circle
Since the radius is 0, the equation of the circle passing through (-4,-4) and tangent to the line 2x-3y+9=0 at (-3,1) is:
(x + 4)^2 + (y + 4)^2 = 0

To find the equation of a circle passing through a given point and tangent to a given line, we need to determine the radius and the center of the circle.

Step 1: Find the equation of the perpendicular bisector of the line segment joining the given point (-4,-4) and the center of the circle.

The center of the circle lies on the perpendicular bisector of the line segment joining (-4,-4) and the tangency point (-3,1). The equation of the line passing through these two points can be found using the midpoint formula:

Midpoint of segment = ( (x₁ + x₂) / 2, (y₁ + y₂) / 2 )

Midpoint = ( (−4 + −3) / 2, (−4 + 1) / 2 ) = (−3.5, −1.5)

Using the point-slope form of a line, where the slope of a perpendicular line is the negative reciprocal of the original line, we can find the equation of the perpendicular bisector:

Slope of the original line = (coefficient of x) / (coefficient of y) = 2 / -3

Slope of the perpendicular bisector = -1 / (2 / -3) = 3/2

Using the point-slope form and substituting the midpoint (−3.5, −1.5), we get:

y - (-1.5) = (3/2)(x - (-3.5))
y + 1.5 = (3/2)(x + 3.5)

Simplifying the equation, we get:

2y + 3 = 3x + 10.5
3x - 2y - 7.5 = 0
6x - 4y - 15 = 0
2x - y - 5/2 = 0 (Dividing by 3 to simplify)

Therefore, the equation of the perpendicular bisector is 2x - y - 5/2 = 0.

Step 2: Find the intersection point of the perpendicular bisector and the given line 2x - 3y + 9 = 0.

To find the intersection point, we can solve the system of equations formed by the perpendicular bisector and the given line. Substituting the equation of the perpendicular bisector into the given line equation, we have:

2x - 3(2x - 5/2) + 9 = 0

Simplifying, we get:

2x - 6x + 15/2 + 9 = 0
-4x + 33/2 = 0
-4x = -33/2
x = (33/2) / 4
x = 33/8

Substituting this value of x into the equation of the given line, we have:

2(33/8) - 3y + 9 = 0
33/4 - 3y + 9 = 0
-3y + 33/4 + 36/4 = 0
-3y + 69/4 = 0
-3y = -69/4
y = (69/4) / 3
y = 23/4

Therefore, the intersection point of the perpendicular bisector and the given line is (33/8, 23/4).

Step 3: Find the distance between the center of the circle and the given point (-4,-4).

The distance between two points (x₁, y₁) and (x₂, y₂) is given by the distance formula:

Distance = √((x₂ - x₁)² + (y₂ - y₁)²)

Distance = √((-4 - (-4))² + (-4 - (-4))²)
Distance = √((0)² + (0)²)
Distance = √(0 + 0)
Distance = √0
Distance = 0

Therefore, the distance between the center of the circle and the given point (-4,-4) is 0.

Step 4: Determine the equation of the circle.

Since the distance between the center of the circle and the given point is 0, the circle is centered at (-4, -4).

The equation of a circle with center (h, k) and radius r is:

(x - h)² + (y - k)² = r²

Substituting the center (-4, -4), we get:

(x - (-4))² + (y - (-4))² = r²
(x + 4)² + (y + 4)² = r²

We still need to find the radius of the circle.

Step 5: Find the distance between the center of the circle and the intersection point of the perpendicular bisector and the given line.

The distance between two points (x₁, y₁) and (x₂, y₂) is given by the distance formula:

Distance = √((x₂ - x₁)² + (y₂ - y₁)²)

Distance = √((33/8 - (-4))² + (23/4 - (-4))²)
Distance = √((33/8 + 32/8)² + (23/4 + 16/4)²)
Distance = √(65/8)² + (39/4)²)
Distance = √(4225/64) + (1521/16)
Distance = √(4225 + 97216) / 64
Distance = √(101441) / 64

Therefore, the distance between the center of the circle and the intersection point is √(101441) / 64.

Since the circle is tangent to the line 2x - 3y + 9 = 0 at the point (-3, 1), the distance between the center of the circle and the intersection point of the perpendicular bisector and the given line is equal to the radius.

Therefore, the radius of the circle is √(101441) / 64.

Substituting the center (-4, -4) and the radius √(101441) / 64 into the equation of the circle, we get:

(x + 4)² + (y + 4)² = (√(101441) / 64)²

Simplifying further, we have:

(x + 4)² + (y + 4)² = 101441 / 4096

Therefore, the equation of the circle passing through (-4, -4) and tangent to the line 2x - 3y + 9 = 0 at (-3, 1) is (x + 4)² + (y + 4)² = 101441 / 4096.

The slope of the given line is 2/3

So, the radius through (-3,1) is perpendicular, with slope -3/2.
So, the line through the center and (-3,1) is

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

Now, the line through the two points forms a chord of the circle, and its slope is 5. So, the radius perpendicular to that point has slope -1/5 and passes through the midpoint of the chord at (-7/2,-3/2).

y+3/2 = -1/5 (x+7/2)

So, now we have two lines which intersect at (-1,-2), the center of the circle. So, the circle is

(x+1)^2 + (y+2)^2 = r^2

The distance from (-1,-2) to (-4,-4) or (-3,1) is √13, so our circle is

(x+1)^2 + (y+2)^2 = 13