Answer
$2\sqrt 2y = 9 -x$
Work Step by Step
To write equation of tangent to circle $x^2 +y^2 = 9$ at the point $(1, 2\sqrt 2)$
Let us find the slope of radius joining centre(0,0) and given point
$m = \frac{2\sqrt 2}{1}$
Tangent is perpendicular to the radius, so its slope can be found
Slope of tangent = $\frac{-1}{m} = \frac{-1}{2\sqrt 2}$
Now we know slope to be $\frac{-1}{2\sqrt 2}$ and one point on tangent $(1, 2\sqrt 2)$
So equation of tangent will be
$y- 2\sqrt 2 = \frac{-1}{2\sqrt 2}(x-1)$
=>$2\sqrt 2y = 9 -x$
So equation of the tangent is $2\sqrt 2y = 9 -x$
