Chapter 9.4, Problem 13MCQ

s

To determine

To find: the centre, radius and the graph of the circle.

Answer to Problem 13MCQ

The centre and radius of the circle are $\left(1,\text{}0\right)$ and radius $3\text{units}$

Explanation of Solution

Given information:

Given the equation of the circle is ${\left(x-1\right)}^2+y^2=9$ .

The standard form of the circle is

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\mathrm{...}\left(1\right)$

The given equation is

  $\begin{array}{l}{\left(x-1\right)}^2+y^2=9\\ {\left(x-1\right)}^2+{\left(y-0\right)}^2=3^2\end{array}$

Compare the equations,

The values of the centre and radius are,

  $h=1,k=0$ and $r=3$

To calculate the points that lie at the circle as,

To put the value of $x=1$ in the given equation

  $\begin{array}{l}{\left(x-1\right)}^2+y^2=9\\ {\left(1-1\right)}^2+y^2=9\\ 0+y^2=9\\ y=\pm 3\end{array}$

To put the value of $y=0$ in the given equation

  $\begin{array}{l}{\left(x-1\right)}^2+0^2=9\\ {\left(x-1\right)}^2=9\\ x-1=3\\ x=\pm 4\end{array}$

The value $x$ of the centre is,

  $x=-2,4$

Thus, the points of the circle are $\left(1,3\right),\left(1,-3\right),\left(2,0\right),\left(4,0\right)$

Hence, the centre and radius of the circle are $\left(1,0\right)$ and radius $3\text{units}$ .

Consider the following graph of the circle,