Find the equation of a circle with center (4, 3) touching the circle `x^2+y^2=1`

Explanation

Let the circle be `x^(2)+y^(2)-8x-6y+k=0` touching the circle `x^(2)+y^(2)=1`. Then the equation of the common tangent at point of contact is `S_(1)-S_(2)=0` or `8x+6y-1-k=0`.
This is a tangent to the circle `x^(2)+y^(2)=1`. Therefore,
`+-1=(k+1)/(sqrt(8^(2)+6^(2)))`
or `k+1=+-10`
i.e., `k= -11` or 9
Hnece, he circles are `x^(2)+y^(2)-8x-6y+9=0` and `x^(2)+y^(2)-8x-6y-11=0`.
Alternative method `:`
The given circle is `x^(2)+y^(2)=1`, which has center `C_(1)(0,0)` and radius `r_(1)=1`. The required circle has center `C_(2)(4,3)` and radius `r_(2)`.
If the circles are touching externally, then
`r_(1)+r_(2)=C_(1)C_(2)`
or `r_(2)=5-1=4`
If the circles are touching internally , then
`r_(2)-r_(1)=C_(1)C_(2)` `( :'(4,3)` lies outside circle `x^(2)+y^(2)=1))`
or `r_(2)=6`
Thus, the required circles are
`(x-4)^(2)+(y-3)^(2)=16`
or `(x-4)^(2)+(y-3)^(2)=36`