SOLUTION: Find the equation of the circle tangent to {{{2x-3y+6 = 0}}} at (-3,0) and contains the point (-3,-6). Thank you!

The center of this circle lies on the straight line perpendicular to the given line and passing through the point (-3,0).


From the other side, the center of this circle lies on the perpendicular bisector to the segment connecting the points (-3,0) and (-3,6).


Notice that this segment, connecting the points (-3,0) and (-3,6), is VERTICAL and its middle point is (-3,3).


So, that perpendicular bisector is simply the horizontal line y = 3.



Perpendicular to the given line has an equation

    3x + 2y = c,      (1)

where "c" is unknown constant.


Since this perpendicular line (1) passes through the point (-3,0), we have

   3*(-3) + 2*0 = c = -9.


Thus, the center of the circle lies on the line

    3x + 2y = -9


and on the line  y =3  at the same time,  which gives you

    3x + 2*3 = -9,

    3x       = -9 - 6 = -15,

     x                = -15/3 = -5.


Thus the center of the circle is the point (x,y) = (-5,3).



Since the distance from the center of the circle to the point (-3,0) is

     =  = ,

the radius of the circle is .



Then the equation of the circle in the standard form is

     +  = ,

or, equivalently,

     +  = 13.      ANSWER