Question

Let $v_0=\left(a,b\right)$and let $v=\left(x,y\right)$Describe the sets of points in $R^2$ satisfying the following properties:

$$\begin{gathered} \left(a\right).\left|v\right|=4 \\ \left(b\right).\left|v\right|\le 4 \\ \left(c\right).\left|v-v_0\right|=4 \end{gathered}$$

Short answer

(a) The set of points are $(0,4),(0,-4),(4,0)\mathrm{and}(-4,0)$etc.

(b) The set of values of $\left(x,y\right)$, which satisfy the relation $x^2+y^2\le 16$will be the required set of points.

The set of points are role="math" localid="1663552322576" $(0,1),(0,2),(0,3),(1,1),(1,2),(1,3),(2,1),(2,2),(3,1),(0,4),(0,-4),(2,3)\mathrm{etc}.$

(c) The set of values of $(x,y)\mathrm{and}(a,b)$, which satisfy the relation ${(x-a)}^2+{(y-b)}^2=16$will be the required set of points.

The set of points $(x,y)$are role="math" localid="1663552768845" $(5,1),(4,1)\mathrm{and}(1,5)\mathrm{etc}.$

The set of points$(a,b)$arerole="math" localid="1663552801181" $(1,1),(0,1)\mathrm{etc}$.

Step-by-step solution

Part (a)

Given that, $v=(x,y)$

role="math" localid="1663552119670" $\left|v\right|=\sqrt{x^2+y^2}$

role="math" localid="1663552143759" $$\begin{gathered} \sqrt{x^2+y^2}=4 \\ {x}^2+y^2=16(\mathrm{Squaring}\mathrm{both}\mathrm{side}) \end{gathered}$$

When $x=0$,

$$\begin{gathered} y=\sqrt{16} \\ y=\pm 4 \end{gathered}$$

When $y=0$,

$$\begin{gathered} x=\sqrt{16} \\ x=\pm 4 \end{gathered}$$

Thus, the set of points are$(0,4),(0,-4),(4,0)\mathrm{and}(-4,0)$etc.

Part (b)

Given that, $v=(x,y)$

$\left|v\right|=\sqrt{x^2+y^2}$

$$\begin{gathered} \sqrt{x^2+y^2}\le 4 \\ {x}^2+y^2\le 16(\mathrm{Squaring}\mathrm{both}\mathrm{side}) \end{gathered}$$

All the points that satisfy the above relationship, will be the required sets of points.

The set of points are,

$(0,1),(0,2),(0,3),(1,1),(1,2),(1,3),(2,1),(2,2),(3,1),(0,4),(0,-4),(2,3)\mathrm{etc}.$

Part (c)

Given that,

$$\begin{gathered} v=(x,y)\mathrm{and}{v}_0=(a,b) \\ v-v_0=\left\{(x-a),(y-b)\right\} \\ \left|v-v_0\right|=\sqrt{{(x-a)}^2+{(y-b)}^2} \\ \sqrt{{(x-a)}^2+{(y-b)}^2}=4\left(\mathrm{Given}\right) \\ {(x-a)}^2+{(y-b)}^2=16\left(\mathrm{Squaring}\mathrm{both}\mathrm{side}\right) \end{gathered}$$

All the values of $(x,y)\mathrm{and}(a,b)$that satisfy the above relationship will be the required sets of points.

The values of $(x,y)$are$(5,1),(4,1)\mathrm{and}(1,5)\mathrm{etc}.$

The values of$(a,b)$are$(1,1),(0,1)\mathrm{etc}$.