Two circles each of radius 5 units touch each other at ( 1 , 2 ) and 4 x + 3 y = 10 is their common tangent.…

The figure below represents the two circles with the common tangent.

$\text{ The slope of the common tangent, }$

$m=-\frac{4}{3}$

$\text{ The slope of the line perpendicular to tangent is, }$

$m^{\text{'}}=\tan \theta =\frac{3}{4}$

$\text{ Therefore, }\sin \theta =\frac{3}{5},\cos \theta =\frac{4}{5}$

$\text{ Now, }$

$x=\left(\pm 5\times \frac{4}{5}+1\right),y=\left(\pm 5\times \frac{3}{5}+2\right)$

$x=(5,-3), y=(5,-1)$

$\text{ The coordinates of }{C}_1\left(5,5\right)\text{ and }{C}_2\left(-3,-1\right)$

$\text{ The equations of the required circles is, }$

$(x-5{)}^2+(y-5{)}^2=5^2$

$x^2+y^2-10x-10y+25=0$

$\text{ And }$

$(x+3{)}^2+(y+1{)}^2=5^2$

$x^2+y^2+6x+2y-15=0$