Are you smarter than me? 12

Let the centers of the circles to the left and right of $4x + 3y = 10$ be $P(a,b)$ and $Q(c,d)$ , respectively.

Since the tangent line has slope $-\frac{4}{3}$ the line $PQ$ lies on has slope $\frac{3}{4}$ . Now the point of tangency of the two circles will be the midpoint of $PQ$ , so

$\frac{a + c}{2} = 1 \Longrightarrow c = 2 - a$ and

$\frac{b + d}{2} = 2 \Longrightarrow d = 4 - b$ .

Next, since $PQ$ has slope $\frac{3}{4}$ and length $10$ , $PQ$ serves as the hypotenuse of a $6/8/10$ right triangle, with

$c - a = 8 \Longrightarrow c = a + 8$ and

$d - b = 6 \Longrightarrow d = b + 6$ .

Comparing these two sets of results, we have that

$2 - a = a + 8 \Longrightarrow a = -3$ and

$4 - b = b + 6 \Longrightarrow b = -1$ .

This in turn gives us $c = -3 + 8 = 5$ and $d = -1 + 6 = 5$ .

Thus the centers of the two circles are $\boxed{(5,5), (-3,-1)}$ .

Alternately, we could use vectors. The unit vector representing the line on which $PQ$ lies is $u = \dfrac{<4,3>}{\sqrt{3^{2} + 4^{2}}} = <\frac{4}{5},\frac{3}{5}>$ , and so

$(a,b) = (1,2) - 5u = (-3,-1)$ and $(c,d) = (1,2) + 5u = (5,5)$ .

Dhaba marili...........