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Explanation

The given circles are
`x^(2)+y^(2)-4x-2y+4=0` (1)
and `x^(2)+y^(2)-12x-8y+36=0` (2)
with centers `C_(1)(2,1)` and `C_(2)(6,4)` and radii 1 and 4 , respectively. Also, `C_(1)C_(2)=5`.
As `r_(1)+r_(2)=C_(1)C_(2)`, the two circles touch each other externally at P.

Clearly, P divides `C_(1)C_(2) ` in the ratio `1:4``,
Therefore, the coordinates of P are
`((1xx6+4xx2)/(1+4),(1xx4+4xx1)/(1+4))-=((14)/(5),(8)/(5))`
Let AB and CD be two common tangents of the given circles, meeting each other at T.
Then T divides `C_(1)C_(2)` externally in the ratio `1:4`. Hence,
`T-= ((1 xx 6-4xx2)/(1-4),(1xx4-4xx1)/(1-4))-=((2)/(3),0)`
Let m be the slope of the tangent. Then the equation of tangent through `(2//3,0)` is
`y-0=m(x-(2)/(3))`
or `y-mx+(2)/(3)m=0`
Now, the length of perpendicular from (2,1) to the above tangent is the radius of the circle. Therefore,
`|(1-2m+(2)/(3)m)/(sqrt(m^(2))+1)|=1`
or `(3-4m)^(2)=9(m^(2)+1)`
or `9-24m+16m^(2)=9m^(2)+9`
or `7m^(2)-24m=0`
or `m=0,(24)/(7)`
Thus, the equations of the tangents are `y=0` and `7y-24x+16=0`.