Find the equation of the circle tangent to the line 2x - 3y + 2 = 0 at (5,4) and with radius equal to √10

Answer 1

To find the equation of the circle, we first need to find the coordinates of its center.

Since the circle is tangent to the line, the center of the circle must lie on the perpendicular line that passes through the point of tangency (5, 4).

The given line has a slope of 2/3, so the perpendicular line will have a slope equal to the negative reciprocal of 2/3, which is -3/2.

Using the point-slope form of a line, the equation of the perpendicular line passing through (5, 4) is:
y - 4 = -3/2(x - 5)

Simplifying, we get:
y - 4 = -3/2x + 15/2
y = -3/2x + 15/2 + 4
y = -3/2x + 23/2

To find the coordinates of the center of the circle, we solve the system of equations consisting of the given line and the perpendicular line:
2x - 3y + 2 = 0
y = -3/2x + 23/2

Substituting -3/2x + 23/2 for y in the first equation, we get:
2x - 3(-3/2x + 23/2) + 2 = 0
2x + 9/2x - 69/2 + 2 = 0
4x + 9x - 69 + 4 = 0
13x - 65 = 0
13x = 65
x = 5

Substituting x = 5 into y = -3/2x + 23/2, we get:
y = -3/2(5) + 23/2
y = -15/2 + 23/2
y = 8/2
y = 4

Therefore, the center of the circle is (5, 4).

Since the radius of the circle is √10, the equation of the circle is:
(x - 5)^2 + (y - 4)^2 = (√10)^2
(x - 5)^2 + (y - 4)^2 = 10

Answer 2

so wrong!
2x - 3y + 2 = 0 is never tangent to the circle
(x - 5)^2 + (y - 4)^2 = 10

The slope of the radius from the center (h,k) of the circle to the point (5,4) is -3/2 So, if (h,k) = (5-2z,4+3z), then we have
√13 z = √10
z = √(10/13)
so the center is at (5-2√(10/13) , 4+3√(10/13))
That makes the equation of the circle
(x - (5-2√(10/13)))^2 + (y - (4+3√(10/13)))^2 = 10

Answer 3

Apologies for the mistake made in the previous response.

The equation of the circle tangent to the line 2x - 3y + 2 = 0 at (5,4) and with a radius equal to √10 is actually:

(x - (5-2√(10/13)))^2 + (y - (4+3√(10/13)))^2 = 10