In this problem, we are asked to find the inverse of the point (2,-3) with respect to the circle x^2+y^2-6x-4y-12=0.


To find the inverse of a point with respect to a circle, we follow the following steps:

1. Find the equation of the given circle.


2. Find the equation of the line passing through the given point and the center of the circle.


3. Find the intersection points of this line with the circle.


4. The inverse of the given point with respect to the circle is the point which is diametrically opposite to the given point with respect to the circle.

Find the equation of the given circle.
The given circle is x^2+y^2-6x-4y-12=0.
We can write the equation of the circle in the standard form as follows:
(x-3)^2+(y-2)^2=25
Therefore, the center of the circle is (3,2) and the radius is 5.


Find the equation of the line passing through the given point and the center of the circle.
The midpoint of the line segment joining the given point (2,-3) and the center of the circle (3,2) is ((2+3)/2,(-3+2)/2) = (2.5,-0.5).
Therefore, the equation of the line passing through the given point and the center of the circle is given by:
y-(-0.5) = (2-(-3))/(3-2.5)*(x-2.5)
Simplifying, we get:
y = 0.4x - 1.3


Find the intersection points of this line with the circle.
Substituting y = 0.4x - 1.3 in the equation of the circle, we get:
(x-3)^2 + (0.4x-2.5)^2 = 25
Simplifying, we get a quadratic equation in x:
1.16x^2 - 12.8x + 33.04 = 0
Solving this quadratic equation, we get two values of x:
x = 1.72, 4.46
Substituting these values in y = 0.4x - 1.3, we get the corresponding values of y:
y = -0.12, 1.98
Therefore, the two intersection points of the line with the circle are (1.72,-0.12) and (4.46,1.98).


The inverse of the given point (2,-3) with respect to the circle is the point which is diametrically opposite to the given point with respect to the circle.
The midpoint of the line segment joining the given point (2,-3) and the intersection point (1.72,-0.12) is ((2+1.72)/2,(-3-0.12)/2) = (1.86