Hello again ivyrianne,
Let's let the equation of the circle be:
$$(x-h)^2+(y-k)^2=r^2$$
We have 3 parameters to determine: $h,k,r$.
In order for the center of the circle to lie on the line $4x+3y=2$, we require:
(1) $$4h+3k=2$$
Now, let's rewrite the tangent lines in slope-intercept form:
a) $$y=-x-4$$
b) $$y=7x+4$$
If we have a point $\left(x_0,y_0 \right)$ and a line $y=mx+b$, then the perpendicular distance $d$ from the point to the line is given by:
$$d=\frac{\left|mx_0+b-y_0 \right|}{\sqrt{m^2+1}}$$
You may find derivations of this formula here:
http://www.mathhelpboards.com/f49/finding-distance-between-point-line-2952/
Using this formula for $d=r$ and the two tangent lines, we find that we require:
$$r=\frac{|-h-4-k|}{\sqrt{2}}=\frac{|7h+4-k|}{5\sqrt{2}}$$
Using the definition $$|x|\equiv\sqrt{x^2}$$, we may square through and write:
$$\left(5(h+k+4) \right)^2=(7h-k+4)^2$$
Solving (1) for $k$, we find:
$$k=\frac{2-4h}{3}$$
Substituting, there results:
$$\left(5\left(h+\frac{2-4h}{3}+4 \right) \right)^2=\left(7h-\frac{2-4h}{3}+4 \right)^2$$
Combining like terms, we find:
$$\left(\frac{5}{3}(h-14) \right)^2=\left(\frac{5}{3}(5h+2) \right)^2$$
Hence, we must have:
$$(h-14)^2-(5h+2)^2=0$$
$$(6h-12)(-4h-16)=0$$
$$-24(h-2)(h+4)=0$$
$$h=-4,\,2$$
Case 1: $$h=-4$$
$$k=\frac{2-4(-4)}{3}=6$$
$$r=\frac{\left|-(-4)-4-6 \right|}{\sqrt{2}}=3\sqrt{2}$$
Case 2: $$h=2$$
$$k=\frac{2-4(2)}{3}=-2$$
$$r=\frac{\left|-2-4-(-2) \right|}{\sqrt{2}}=2\sqrt{2}$$
Thus, the two possible circles are:
$$\left(x+4 \right)^2+\left(y-6 \right)^2=18$$
$$\left(x-2 \right)^2+\left(y+2 \right)^2=8$$
Here is a plot of the three lines and the two circles:
https://www.physicsforums.com/attachments/1008._xfImport
