Best Answer:
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

Collect terms:

(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.

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