Chapter 1, Problem 129RE

To determine

To calculate:The equation for the circle and line in the given figure.

Answer to Problem 129RE

The equation for the circle and line in the figure are $x^2+y^2=169$ and $5x-12y+169=0$ respectively.

Explanation of Solution

Given information:

  

The point $P\left(-5,12\right)$ lying on the tangent to the circle with center $\left(h,k\right)=\left(0,0\right)$

Formula used:

For a given circle with center $\left(h,k\right)$ and radius $r$ the equation is given as:

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

This is referred to as the Standard form for the equation of a given circle.

The distance $D$ between two points say $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given as:

  $D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Slope m of the line passing through two points in general say $P_1=\left(x_1,y_1\right)$ and $P_2=\left(x_2,y_2\right)$ is:

  $m=\frac{y_2-y_1}{x_2-x_1}$

Slope-intercept equation for a given line which has slope as $m$ and $y$ −intercept as $b$ is:

  $y=mx+b$

Two-intercept equation for a given line which has $x$ -intercept as $a$ and $y$ −intercept as $b$ is:

  $\frac{x}{a}+\frac{y}{b}=1$

When two lines are perpendicular then the product of their slopes is zero that is $m_1.m_2=-1$

When two lines are parallel then their slope are equal that is $m_1=m_2$

Calculation:

From the given figure it is clear that center of the circle is $\left(h,k\right)=\left(0,0\right)$

Recall for a given circle with center $\left(h,k\right)$ and radius $r$ the equation is given as:

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Therefore, replacing the values we get:

  ${\left(x-0\right)}^2+{\left(y-0\right)}^2=r^2$

  $x^2+y^2=r^2$ $\left(1\right)$

Now to find radius $r$ which is the distance from the center $\left(h,k\right)=\left(0,0\right)$ to the point $P\left(-5,12\right)$

Recall, the distance $D$ between two points say $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given as:

  $D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Before applying distance formula the following must be known:

  $\begin{array}{l}{x}_1=0\\ y_1=0\\ x_2=-5\\ y_2=12\end{array}$

Therefore,

  $\begin{array}{l}D=r=\sqrt{{\left(-5-0\right)}^2+{\left(12-0\right)}^2}\\ =\sqrt{25+144}\\ =\sqrt{169}\\ =13\end{array}$

Hence radius $r=13$

Put this value in $\left(1\right)$

Therefore, the required equation for circle is:

  $\begin{array}{l}{x}^2+y^2={13}^2\\ x^2+y^2=169\end{array}$

Hence, the equation for the given circle is $x^2+y^2=169$

Now for the equation of line:

Recall, slope m of the line passing through two points in general say $P_1=\left(x_1,y_1\right)$ and $P_2=\left(x_2,y_2\right)$ is:

  $m=\frac{y_2-y_1}{x_2-x_1}$

As it is known:

  $\begin{array}{l}{x}_1=0\\ y_1=0\\ x_2=-5\\ y_2=12\end{array}$

Therefore, slope of radius for above values is:

  $\begin{array}{l}{m}_1=\frac{12-0}{-5-0}\\ m_1=\frac{-12}{5}\end{array}$

Now as the tangent is perpendicular to the circle, therefore the product of their slopes is $-1$

Slope of line is $m_2=\frac{5}{12}$

And it passes through $P\left(-5,12\right)$ that is $\begin{array}{l}x=-5\\ y=12\end{array}$

Put these values in the general equation for the line which is $y=m_{2}x+b$ $\left(2\right)$

Which gives:

  $\begin{array}{l}12=\frac{5}{12}\left(-5\right)+b\\ b=12+\frac{25}{12}\\ b=\frac{144+25}{12}\\ b=\frac{169}{12}\end{array}$

Now put $m_2=\frac{5}{12}$ and $b=\frac{169}{12}$ in $\left(2\right)$ to get:

  $\begin{array}{l}y=\frac{5}{12}x+\frac{169}{12}\\ 12y=5x+169\\ 5x-12y+169=0\end{array}$

Thus, required equation of line is $5x-12y+169=0$