Chapter 1, Problem 130RE

To determine

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

Answer to Problem 130RE

The equation for the circle and line in the figure are $x^2+y^2-10x-10y+25=0$ and $4x-3y+35=0$ respectively.

Explanation of Solution

Given information:

  

The point $P\left(8,1\right)$ lying on the tangent to the circle with center $\left(h,k\right)=\left(5,5\right)$

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 general equation for the circle is:

  $x^2+y^2+2gx+2fy+c=0$

Where $\left(-g,-f\right)$ denotes the center of the circle and radius is given by $r=\sqrt{g^2+f^2-c}$

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 $\left(h,k\right)=\left(5,5\right)$

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:

  ${\left(x-5\right)}^2+{\left(y-5\right)}^2=r^2$ $\left(1\right)$

Now to find radius $r$ which is the distance from the center $\left(h,k\right)=\left(5,5\right)$ to the point $P\left(8,1\right)$

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:

  $\begin{array}{l}{x}_1=5\\ y_1=5\\ x_2=8\\ y_2=1\end{array}$

Therefore,

  $\begin{array}{l}D=r=\sqrt{{\left(8-5\right)}^2+{\left(1-5\right)}^2}\\ =\sqrt{9+16}\\ =\sqrt{25}\\ =5\end{array}$

Hence radius $r=5$

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

  $\begin{array}{l}{\left(x-5\right)}^2+{\left(y-5\right)}^2=5^2\\ {\left(x-5\right)}^2+{\left(y-5\right)}^2=25\end{array}$

Recall the identities:

  $\begin{array}{l}{\left(a+b\right)}^2=a^2+2ab+b^2\\ {\left(a-b\right)}^2=a^2-2ab+b^2\end{array}$

Use this to get:

  $\begin{array}{l}{x}^2+25-10x+y^2+25-10y=25\\ x^2+y^2-10x-10y+25=0\end{array}$

Hence, the equation for the given circle is $x^2+y^2-10x-10y+25=0$

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: $P_1=\left(5,5\right)$ and $P_2=\left(8,1\right)$

  $\begin{array}{l}{x}_1=5\\ y_1=5\\ x_2=8\\ y_2=1\end{array}$

Therefore, slope of line is:

  $\begin{array}{l}m=\frac{1-5}{8-5}\\ m=\frac{-4}{3}\end{array}$

Slope of line is $m=\frac{-4}{3}$

And it passes through $P_1=\left(5,5\right)$ that is $\begin{array}{l}x=5\\ y=5\end{array}$

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

Which gives:

  $\begin{array}{l}5=\frac{-4}{3}\left(5\right)+b\\ b=5+\frac{20}{3}\\ b=\frac{15+20}{3}\\ b=\frac{35}{3}\end{array}$

Now put $m=\frac{-4}{3}$ and $b=\frac{35}{3}$ in $\left(2\right)$ to get:

  $\begin{array}{l}y=\frac{-4}{3}x+\frac{35}{3}\\ 3y=-4x+35\\ 4x-3y+35=0\end{array}$

Thus, required equation of line is $4x-3y+35=0$