Chapter 1.4, Problem 54AYU

To determine

To find: an equation of the tangent line.

Answer to Problem 54AYU

Therefore, the equation of the tangent line is $\sqrt{2}x+4y+12-11\sqrt{2}=0$ .

Explanation of Solution

Given:

U $x^2+y^2-4x+6y+4=0$

  $\left(3,2\sqrt{2}-3\right)$ .

Since the line containing the center and the tangent line are perpendicular, the product of their slopes will be $-1$ . Let $m_1$ represent the slope of the line containing the center and $m_2$ represent the slope of the tangent line.

Group the terms containing $x$ and the terms containing $y$ and put the constant on the right side of the equation.

  $\left(x^2-4x\right)+\left(y^2+6y\right)=-4$

Complete the square of each expression in parentheses. The numbers added to the left-hand side must be added to the right-hand side also.

Factor the expression.

  $\begin{array}{l}\left(x^2-4x+4\right)+\left(y^2+6y+9\right)=9\\ {\left(x-2\right)}^2+{\left(y+3\right)}^2=3^2\end{array}$

The standard from of an equation of a circle with radius $r$ and center at is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ . Since the circle ${\left(x-2\right)}^2+{\left(y+3\right)}^2=3^2$ is in this from, its center is at $\left(2,-3\right)$ , and the radius is $3$ .

It is given that $\left(3,2\sqrt{2}-3\right)$ is a point on the circle. The slope of the line passing through $\left(2,-3\right)$ and the can be found using.

Substitute for the values

  $\begin{array}{l}{m}_1=\frac{2\sqrt{2}-3-\left(-3\right)}{3-2}\\ =\frac{2\sqrt{2}-3+3}{1}\\ =2\sqrt{2}\end{array}$

Now, find $m_2$ . The product of $m_1$ and $m_2$ is $-1$ . Thus

  $\begin{array}{l}2\sqrt{2}.m_2=-1\\ m_2=-\frac{1}{2\sqrt{2}}\end{array}$

The slope $m_2$ is $-\frac{1}{2\sqrt{2}}$ .

The tangent also has the point $\left(3,2\sqrt{2}-3\right)$ . The point-slope form of the equation f the tangent can be written using.

  $\left(y-\left(2\sqrt{2}-3\right)\right)=-\frac{1}{2\sqrt{2}}\left(x-3\right)$

Multiply the numerator and denominator of right-hand side of the equation by $\sqrt{2}$ .

  $\begin{array}{l}\left(y-2\sqrt{2}+3\right)=\frac{-1.\sqrt{2}}{2\sqrt{2}.\sqrt{2}}\left(x-3\right)\\ \left(y-2\sqrt{2}+3\right)=\frac{\sqrt{2}}{4}\left(x-3\right)\end{array}$

Simplify

  $\begin{array}{l}4\left(y-2\sqrt{2}+3\right)=\left[-\frac{\sqrt{2}}{4}\left(x-3\right)\right]4\\ 4y-8\sqrt{2}+12=-\sqrt{2}\left(x-3\right)\\ 4y-8\sqrt{2}+12=-\sqrt{2}x+3\sqrt{2}\end{array}$

Subtract $-\sqrt{2}x+3\sqrt{2}$ from both sides of the equation.

  $\begin{array}{l}4y-8\sqrt{2}+12-\left(-\sqrt{2}x+3\sqrt{2}\right)=-\sqrt{2}x+3\sqrt{2}-\left(-\sqrt{2}x+3\sqrt{2}\right)\\ 4y-8\sqrt{2}+12+\sqrt{2}x-3\sqrt{2}=0\\ 4y+\sqrt{2}x+12-11\sqrt{2}=0\end{array}$

Therefore, the equation of the tangent line is $\sqrt{2}x+4y+12-11\sqrt{2}=0$ .

Conclusion:

Therefore, the equation of the tangent line is $\sqrt{2}x+4y+12-11\sqrt{2}=0$ .