Chapter 1.4, Problem 53AYU

To determine

To find: The equation of the tangent line

Answer to Problem 53AYU

The equation of the tangent line is $\sqrt{2}x-4y-9\sqrt{2}=0$

Explanation of Solution

Given:

  $x^2+y^2=9$

  $\left(1,2\sqrt{2}\right)$

Calculation:

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

The standard form of an equation of a circle with its center at the origin and the radius $r$ is

  $x^2+y^2=r^2$

Since the given circle $x^2+y^2=9$ is in this form, its center is the origin $\left(0,0\right)$ and the radius is $\sqrt{9}=3$

  $\left(1,2\sqrt{2}\right)$ is a point on the circle.The slope of the line passing through the origin and the point $\left(1,2\sqrt{2}\right)$ can be found using $m_1=\frac{y_2-y_1}{x_2-x_1}$

  $m_1=\frac{2\sqrt{2}-0}{1-0}$

  $=2\sqrt{2}$

The product of $m_1$ and $m_2$ is $-1$

  $m_1{m}_2=-1$

  $2\sqrt{2}·m_2=-1$

  $m_2=-\frac{1}{2\sqrt{2}}$

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

The tangent also has the point $\left(1,2\sqrt{2}\right)$ .

So, the point-slope form of the equation of the tangent can be written using $\left(y-y_1\right)=m_2\left(x-x_1\right)$

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

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

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

  $\left(y-2\sqrt{2}\right)=-\frac{\sqrt{2}}{4}\left(x-1\right)$

Multiply both side of the equation by 4 to clear the fraction

  $4\left(y-2\sqrt{2}\right)=4\left(\frac{\sqrt{2}}{4}\left(x-1\right)\right)$

  $4\left(y-2\sqrt{2}\right)=-\sqrt{2}\left(x-1\right)$

Remove the parentheses using the distributive property

  $4y-8\sqrt{2}=-\sqrt{2}x+\sqrt{2}$

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

  $4y-8\sqrt{2}+\sqrt{2}x-\sqrt{2}=-\sqrt{2}x+\sqrt{2}+\sqrt{2}x-\sqrt{2}$

  $\sqrt{2}x-4y-9\sqrt{2}=0$

Conclusion:

Hence,the equation of the tangent line is $\sqrt{2}x-4y-9\sqrt{2}=0$