Hello, prasum!
Here's a back-door approach . . .
Two circles each of radius 5 units touch each other at (1,2).
If the equation of common tangent is $\displaystyle 4x+3y\,=\,10$,
find the equation of two circles. Code:
*
|\ * * *
| \ * *
-4 | \ * *
| \ * *
*- - *
3 \ * P *
\ * ♠ *
\* o | *
* * * \ o |
* ** o | *
* * - - - ◊ - - - * *
* | o ** *
| o \ * * *
* | o *\
* ♣ * \
* Q * \
\
* * \
* * \
* *
* * *
The two circles are tangent at: .$\displaystyle \diamondsuit(1,2)$
The centers of the circles are: .$\displaystyle P\spadesuit \text{ and }Q\clubsuit$
The tangent has slope $\displaystyle -\frac{4}{3}$
The radii, $\displaystyle P\diamondsuit \text{ and }Q\diamondsuit$ are perpendicular to the tangent
. . and have length 5.
From $\displaystyle \diamondsuit(1,2)$ move 4 units right and 3 units up to $\displaystyle P(5,5)$
. . and we find that: .$\displaystyle |P\diamondsuit| = 5$
From $\displaystyle \diamondsuit(1,2)$ move 4 units left and 3 units down to $\displaystyle Q(\text{-}3,\text{-}1)$
. . and we find that: .$\displaystyle |Q\diamondsuit| = 5$
Hence, the centers of the circles are: .$\displaystyle P(5,5) \text{ and }Q(\text{-}3,\text{-}1)$
Therefore, their equations are: .$\displaystyle \begin{Bmatrix}(x-5)^2 + (y-5)^2 &=^& 25 \\ (x+3)^2 + (y+1)^2 &=& 25 \end{Bmatrix}$