Find the equations of tangents to the circle `x^2+y^2-22 x-4y+25=0` which are perpendicular to the line `5x+12 y+8=0`

Answer

`12x-5y+8=0` and `12x-5y=252`

Explanation

Given circle is `x^(2)+y^(2)-22x-4y+25=0`.
Centre is (11,2) and radius is `sqrt(121+4-25)=10`
Equation of line perpendicular to `5x+12y+8=0` is `12x-5y+k=0` .
If this line touches the circle, then distance of the centre of the circle from the line radius.
`:. |(12(11)-5(2)+k)/(sqrt(144+25))|=10`
`implies |k+122|=130`
`implies k=8` or `-252`
Hence equations of tangents are
`12x+5y+8=0` and `12x-5y=252`.