Q.53 - Precalculus Enhanced with Graphing Utilities

Question

The Greek Method The Greek method for finding the equation of the tangent line to a circle uses the fact that at any point on a circle the lines containing the center and the tangent line are perpendicular (see Problem 52). Use this method to find an equation of the tangent line to the circle

x² + y² = 9 at the point $(1 ,2\sqrt{2).}$

Step-by-Step Solution

Answer

An equation of the tangent line to the circle $x^2+y^2=9$ at the point $(1 ,2\sqrt{2})$ is$y-2\sqrt{2}=-\frac{1}{2\sqrt{2}}(x-1) .$

Step 1. Given information

 Given that equation of  a circle $x^2+y^2=9$ and tangent line point$(1 ,2\sqrt{2).}$

 we need to find out  an equation of the tangent line to the circle  at the given point

Step 2. Description of finding the slope

Here is the equation of a circle $x^2+y^2=9$ and its center(h ,k) is (0,0).

Slope of line joining point to the given point is

  $$\begin{gathered} m =\frac{y_2-y_1}{x_{2-}{x}_1} \\ =\frac{2\sqrt{2}}{1} \\ =2\sqrt{2 } \end{gathered}$$

Here stated that at point on a circle the lines containing the center and the tangent line are perpendicular,

thus $$\begin{gathered} m_2 =\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$m_1$}\right. \\ =-\frac{1}{2\sqrt{2}} \end{gathered}$$

Step 3. Description of finding equation of the tangent line to the circle

The equation of a line passing through (x_{1 ,} y₁) and slope m is $$\begin{gathered} (y-y_1)=m(x-x_1) \\ (y-2\sqrt{2})=\frac{-1}{2\sqrt{2}}(x-1) , here line passes through (1,2\sqrt{2}) so (x_1,y_1) is (1,2\sqrt{2}) \end{gathered}$$