Answer
$ 2\sqrt2 y = 11- 6\sqrt2 - x$
Work Step by Step
Given circle $x^2 + y^2 - 4x + 6y + 4 = 0$
To find equation of the tangent at the point $(3, 2\sqrt2 - 3)$
Let us find center of the circle first
Equation of the circle can be rewritten as
$(x-2)^2 + (y+3)^2 = (3)^2$
This center of circle as $(2, -3)$
Since tangent at any point on circle is perpendicular to the line joining center of the circle and that point
Slope of line joining center of the circle and that point $(3, 2\sqrt2 - 3)$
= $\frac{-3-( 2\sqrt2 - 3)}{2-3} = 2\sqrt2$
So slope of the tangent will be = $\frac{-1}{2\sqrt2}$
Equation of tangent through point $(3, 2\sqrt2 - 3)$ and slope $\frac{-1}{2\sqrt2}$ will be
$y- (2\sqrt2 - 3) = \frac{-1}{2\sqrt2}(x-3)$
=>$ 2\sqrt2 y - 8 + 6\sqrt2 = 3-x$
=> $ 2\sqrt2 y = 11- 6\sqrt2 - x$
