Use the Greek method described in Problem 53 to find an equation of the tangent line to the circle

x² + y² - 4x + 6y + 4 = 0 at the point$\left(3,2\sqrt{2}-3\right)$

Here equation of a circle $x^2+y^2-4x+6y+4=0atthepoint(3,2\sqrt{2}-3)isgiven.$

We have to find out the equation of the tangent line to the circle .

To find center , first convert equation in a standard form ,$$\begin{gathered} x^2+y^2-4x+6y+4=0 \\ groupthexcontainingtermandycontainingterm \\ (x^2-4x)+(y^2+6y)=-4,nowcompletethesquareinsidetheparenthesis \\ (x^2-4x+4)+(y^2+6y+9)=-4+4+9 \\ {(x-2)}^2+{(y+3)}^2=9,thisisinthes\tan dardformofequationofacircle \\ hencecenteris(2,-3) \end{gathered}$$

The slope of the line joining center (2,-3) to a given point$(3,2\sqrt{2}-3)$, then role="math" localid="1646304685996" $$\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ =\frac{2\sqrt{2}-3+3}{3-2} \\ =2\sqrt{2} \\ {m}_2=\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$m$}\right.\sin ceitisperpendiculartolinejoiningcenter \\ =\frac{-1}{2\sqrt{2}}thisslopeislinepas\sin gthroughthepoint(3,2\sqrt{2}-3). \end{gathered}$$

The equation of a line passing through (x₁,y₁ ) and slop m is$$\begin{gathered} (y-y_1)=m(x-x_1) \\ (y-(2\sqrt{2}-3\left)\right)=\frac{-1}{2\sqrt{2}}(x-3),here\left(x_{1,}{y}_1\right)is(3,2\sqrt{2}-3)amdm=\frac{-1}{2\sqrt{2}} \\ y-2\sqrt{2}+3=\frac{-1}{2\sqrt{2}}(x-3) \\ (y-2\sqrt{2}+3)\times 2\sqrt{2}=-x+3 \\ 2\sqrt{2}y-8+6\sqrt{2}=-x+3 \\ 2\sqrt{2}y+x=11-6\sqrt{2},thisistheequationofthe\tan gentlinetothecircle. \end{gathered}$$