# For the equation $(x-2)^2+(y-2)^2=4$ find the equation of tangent at $(2,4)$?

$\begin{array}{1 1}(a)\;y=3\\(b)\;y=6\\(c)\;y=4\\(d)\;y=2\end{array}$

We know the tangent for a circle is $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$-------(1)
Now comparing $x^2+y^2+2gx+2fy+c=0$ to $(x-2)^2+(y-2)^2=4$------(2)
Equ(1) can be written as $x^2+y^2-4x-4y+4=0$
Hence $g=-2,f=-2,c=4$
Now putting g & f in (2) we get
$2x+4y-2(x+2)-2(y+4)+4=0$
$2y=8$
$y=4$
Hence (c) is the correct answer.