The figure below represents the two circles with the common tangent.
The slope of the common tangent, m=−
4
3
The slope of the line perpendicular to tangent is, m′=tanθ=
3
4
Therefore, sinθ=
3
5
,cosθ=
4
5
Now,
x−1
4∕5
=
y−1
3∕5
=±5 Now, x=(±5×
4
5
+1),y=(±5×
3
5
+2) x=(5,−3),y=(5,−1) The coordinates of C1(5,5) and C2(−3,−1) The equations of the required circles is, (x−5)2+(y−5)2=52 x2+y2−10x−10y+25=0 And (x+3)2+(y+1)2=52 x2+y2+6x+2y−15=0