Find an equation of the circle tangent to x + y = 3 at (2,1) and with center on 3x - 2y - 6 = 0

y = - x + 3 so

slope of tangent = -1
slope of line from (2,1) to circle center = -1/-1 = 1 because radius of circle is perpendicular to tangent at intersection
so find the radius line to the tangent point (2,1)
y = +1 x + b
1 = 2 + b
b = -1
so that radius line is y = x-1
the center is also on 3 x - 2 y = 6
so
3 x - 2 (x-1) = 6
x = 4
y = 3
center at (4,3)
so form is
(x-4)^2 + (y-3)^2 = r^2
we still need r^2 but we know center at (4,3) and tangent point at (2,1)
distance between is r
r^2 = (1-3)^2 + (2-4)^2
r^2 = 4 + 4 = 8
so
(x-4)^2 + (y-3)^2 = 8
check my arithmetic, I did that fast.

Why did the circle decide to make friends with the line?

Because they both love tangents!

Now, let's get down to business and solve this problem.

The equation of a circle with center (h, k) and radius r is given by (x - h)^2 + (y - k)^2 = r^2.

First, we know that the center of the circle lies on the line 3x - 2y - 6 = 0. So, let's find the coordinates of the center.

We can rewrite the equation of the line as 3x - 2y = 6. To find the x-coordinate of the center, we can equate y to 0 and solve for x: 3x - 2(0) = 6. This gives us x = 2.

Next, to find the y-coordinate of the center, we can equate x to 0 and solve for y: 3(0) - 2y = 6. Solving for y, we get y = -3.

So, the center of the circle is (2, -3).

Now, we know that the circle is tangent to the line x + y = 3 at the point (2, 1). The radius of the circle is the distance from the center to the point of tangency.

Using the distance formula, we can calculate the radius:

r = sqrt((2 - 2)^2 + (1 - (-3))^2)
= sqrt(0 + 16)
= 4

Therefore, the equation of the circle is:

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

Simplifying, we get:

(x - 2)^2 + (y + 3)^2 = 16

So, the equation of the circle tangent to x + y = 3 at (2,1) and with center on 3x - 2y - 6 = 0 is (x - 2)^2 + (y + 3)^2 = 16.

And that's how the circle made friends with the line while staying funny!

To find the equation of the circle tangent to the line x + y = 3 at (2,1) and with its center on the line 3x - 2y - 6 = 0, we can follow these steps:

Step 1: Determine the center of the circle.
Since the center of the circle lies on the line 3x - 2y - 6 = 0, we can rewrite this equation in slope-intercept form:
3x - 2y - 6 = 0
-2y = -3x + 6
y = (3/2)x - 3

The slope-intercept form of a line is y = mx + b, where m is the slope and b is the y-intercept. From this equation, we can see that the slope is (3/2) and the y-intercept is -3.

To find the x-coordinate of the center of the circle, we use the fact that the center lies on the line x + y = 3:
x + y = 3
x + (3/2)x - 3 = 3
(5/2)x = 6
x = 12/5

To find the y-coordinate of the center, we substitute the x-coordinate we just found into the equation y = (3/2)x - 3:
y = (3/2)(12/5) - 3
y = 18/5 - 15/5
y = 3/5

Therefore, the center of the circle is at (12/5, 3/5).

Step 2: Determine the radius of the circle.
Since the circle is tangent to the line x + y = 3 at the point (2,1), the distance between the center of the circle and the point (2,1) is equal to the radius of the circle.

Using the distance formula, the radius, r, is given by:
r = √[(x2 - x1)^2 + (y2 - y1)^2]
r = √[(2 - 12/5)^2 + (1 - 3/5)^2]
r = √[(10/5 - 12/5)^2 + (5/5 - 3/5)^2]
r = √[(-2/5)^2 + (2/5)^2]
r = √[4/25 + 4/25]
r = √[8/25]
r = √(8)/√(25)
r = 2√(2)/5

Therefore, the radius of the circle is 2√(2)/5.

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

Substituting the values we found, we have:
(x - 12/5)^2 + (y - 3/5)^2 = (2√2/5)^2
(x - 12/5)^2 + (y - 3/5)^2 = 8/25

Thus, the equation of the circle tangent to x + y = 3 at (2,1) and with center on 3x - 2y - 6 = 0 is:
(x - 12/5)^2 + (y - 3/5)^2 = 8/25

To find the equation of a circle, we need the coordinates of its center and the radius. We are given that the circle is tangent to the line x + y = 3 and its center lies on the line 3x - 2y - 6 = 0.

Step 1: Find the slope of the line x + y = 3.
To find the slope of the line, we can write it in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.
Rearranging the equation x + y = 3, we have y = -x + 3, which means the slope is -1.

Step 2: Find the perpendicular slope to the line x + y = 3.
The slope of a line perpendicular to another line is the negative reciprocal of its slope. Therefore, the perpendicular slope to -1 is 1.

Step 3: Determine the equation of the line passing through the point (2,1) with a slope of 1.
Using the point-slope form (y - y1 = m(x - x1)), where (x1, y1) is the point on the line, we substitute the values into the equation: y - 1 = 1(x - 2), which simplifies to y = x - 1.

Step 4: Find the intersection point of the lines 3x - 2y - 6 = 0 and y = x - 1.
To find the intersection point, we need to solve the system of equations. Substituting the equation of the line (y = x - 1) into the equation of the line (3x - 2y - 6 = 0), we have 3x - 2(x - 1) - 6 = 0. Simplifying, we get x = 2.

Step 5: Substitute the x-coordinate into the equation of the line to find the y-coordinate.
Using the equation of the line y = x - 1, we substitute x = 2 to find y = 2 - 1 = 1.

Step 6: The coordinates of the center are (2,1).
Since the center of the circle lies on the line 3x - 2y - 6 = 0 and intersects the line x + y = 3 at (2,1), the coordinates of the center are (2,1).

Step 7: Find the radius of the circle.
The distance between the center of the circle (2,1) and any point on the line x + y = 3 will give us the radius. Let's take a point on the line, for example, (0,3).
Using the distance formula, the radius is √[(x2 - x1)^2 + (y2 - y1)^2] = √[(0 - 2)^2 + (3 - 1)^2] = √[(-2)^2 + 2^2] = √[4 + 4] = √8 = 2√2.

Step 8: Write the equation of the circle.
The equation of a circle with center (h, k) and radius r is given by (x - h)^2 + (y - k)^2 = r^2.
Substituting the values we found, the equation of the circle is (x - 2)^2 + (y - 1)^2 = (2√2)^2.

Simplifying, we have:
(x - 2)^2 + (y - 1)^2 = 8