Chapter 13.6, Problem 38WE

a.

To determine

To find:Verify that the point $P(4,-2)$ is on the line $2x-y=10$ and on the circle $x^2+y^2=20$ .

Explanation of Solution

Given information:Point $P(4,-2)$ , line $2x-y=10$ and circle $x^2+y^2=20$ .

Calculation:

Substitute values of x and y from point $P(4,-2)$ in equation of the line $2x-y=10$ :

  $\begin{array}{l}2\times 4-(-2)=10\\ \Rightarrow 8+2=10\\ \Rightarrow 10=10\end{array}$

Since, point $P(4,-2)$ satisfied the equation of line.

So, point $P(4,-2)$ lies on the line $2x-y=10$ .

Substitute values of x and y from point $P(4,-2)$ in equation of the circle $x^2+y^2=20$ :

  $\begin{array}{l}$ 4^2+{(-2)}^2=20$\Rightarrow 16+4=20\\ \Rightarrow 20=20\end{array}$

Since, point $P(4,-2)$ satisfied the equation of circle $x^2+y^2=20$ .

So, point $P(4,-2)$ lies on the circle $x^2+y^2=20$ .

Hence verified.

b.

To determine

To find:Show that the segment joining the center of circle to $P$ is perpendicular to the line.

Explanation of Solution

Given information:Point $P(4,-2)$ , line $2x-y=10$ and circle $x^2+y^2=20$ .

Formula used: Slope of the line $ax+by+c=0$ is $-\frac{a}{b}$ .

Theslope of a line passing through point $(x_1,y_1)$ and $(x_2,y_2)$ is $\frac{y_2-y_1}{x_2-x_1}$ .

Concept used:Two lines are perpendicular if product of their slopes is $-1$ .

i.e. if slopes of lines are $m_1$ and $m_2$ then $m_1{m}_2=-1$ .

Center of circle ${(x-h)}^2+{(y-k)}^2=r^2$ is $(h,k)$

Calculation:

Since, equation of circle is $x^2+y^2=20$ .

So, center of circle is $O(0,0)$

Let slope of segment $OP$ be $m_1$

  $\begin{array}{l}{m}_1=\frac{y_2-y_1}{x_2-x_1}\\ \Rightarrow m_1=\frac{-2-0}{4-0}\\ \Rightarrow m_1=-\frac{1}{2}\end{array}$

Let slope of line $2x-y=10$ be $m_2$

  $m_2=-\frac{a}{b}$

Here, $a=2,b=-1$

  $\begin{array}{l}{m}_2=-\frac{2}{\left(-1\right)}\\ \Rightarrow m_2=2\end{array}$

Now,

  $\begin{array}{l}{m}_1{m}_2=\left(-\frac{1}{2}\right)\times 2\\ \Rightarrow m_1{m}_2=-1\end{array}$

Since, $m_1{m}_2=-1$

Hence, segment joining the center of circle to $P$ is perpendicular to the line.

c.

To determine

To find: Show that the segment joining the center of circle to $P$ is perpendicular to the line.

Answer to Problem 38WE

The line $2x-y=10$ is tangent to the circle $x^2+y^2=20$ .

Explanation of Solution

Given information:Point $P(4,-2)$ , line $2x-y=10$ and circle $x^2+y^2=20$ .

Concept used:A point lies on both line and circle such that the segment joining the center of circle to this point is perpendicular to the line then the line is tangent to the circle at that point.

Calculation:

From the parts (a) and (b):

Since, point $P(4,-2)$ is on the line $2x-y=10$ and on the circle $x^2+y^2=20$ as well assegment joining the center of circle to $P$ is perpendicular to the line.

So, line $2x-y=10$ is tangent to the circle $x^2+y^2=20$ at point $P(4,-2)$ .

Hence, the line $2x-y=10$ is tangent to the circle $x^2+y^2=20$ .