Let's first find the intersection point of the tangent lines:
subtract the LHS of the second eqn from that of the first eqn.
3y + 13 + 3y + 1 = 0
6y = - 14 => y = -7/3
x = (3y - 1)/2 = -4
The slopes of the tangents are -2/3 and 2/3,
so the bisectors of the angles between the tangents are
parallel to the x and y axes; in particular, the center of
the circle must lie either on the line x = -4 or the line y = -7/3.
The point (0,4) lies "above" both tangents, so the circle
specified has its center on the line x = -4. The equation of the
circle is of the form
(x+4)^2 + (y-k)^2 = r^2, [ Equation 0 ]
where we still need to determine k and r.
Since the specifed circle must contain (0,4), we know that
16 + k^2 = r^2. [ Equation A ]
The distance from (-4, k) to the line 2x - 3y - 1 = 0 is r, and
that distance is measured along a line whose slope is -3/2
because it's perpendicular to 2x-3y-1=0. This particular radius
is the hypotenuse of a right triangle whose legs have the ratio
-3/2 and satisfy the Pythagorean Theorem: their lengths are
2 r /sqrt(13) and -3 r /sqrt(13). The coordinates
of the point of tangency are -4 + 2 r / sqrt(13) and
k - 3 r / sqrt(13). The requirement that this point lie on the
line 2x - 3y - 1 = 0 gives us
-8 + 4 r /sqrt(13) - 3k - 9 r /sqrt(13) - 1 = 0 [ Equation B ]
Now, Equations A and B can be simultaneously solved
to obtain k and r:
From Eqn A, k = sqrt( r^2 - 16 ), so Eqn B becomes
-8 + 4 r /sqrt(13) - 3 sqrt( r^2 - 16 ) - 9 r /sqrt(13) - 1 = 0
We need to eliminate the unknown radicand, so we rearrange:
3 sqrt( r^2 - 16 ) = - 9 - 5 r /sqrt(13)
Square both sides:
9 r^2 - 144 = 81 + 90 r sqrt(13) + (25/13) r^2
(92/13) r^2 - 90 r sqrt(13) - 225 = 0 [ Equation C ]
At this point, you should review my work in detail
and correct any arithmetic errors.
Then use the quadratic formula to solve Equation C,
obtaining r. Plug that back into Eqn A to get k.
Plug r and k into [ Equation 0 ] to obtain the
explicit equation of the required circle.