# 7.1 Coordinate geometry, equation of tangent, transformations  (Page 2/2)

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${m}_{f}=\frac{{y}_{1}-{y}_{0}}{{x}_{1}-{x}_{0}}$

The tangent (line $g$ ) is perpendicular to this line. Therefore,

${m}_{f}×{m}_{g}=-1$

So,

${m}_{g}=-\frac{1}{{m}_{f}}$

Now, we know that the tangent passes through $\left({x}_{1},{y}_{1}\right)$ so the equation is given by:

$\begin{array}{ccc}\hfill y-{y}_{1}& =& m\left(x-{x}_{1}\right)\hfill \\ \hfill y-{y}_{1}& =& -\frac{1}{{m}_{f}}\left(x-{x}_{1}\right)\hfill \\ \hfill y-{y}_{1}& =& -\frac{1}{\frac{{y}_{1}-{y}_{0}}{{x}_{1}-{x}_{0}}}\left(x-{x}_{1}\right)\hfill \\ \hfill y-{y}_{1}& =& -\frac{{x}_{1}-{x}_{0}}{{y}_{1}-{y}_{0}}\left(x-{x}_{1}\right)\hfill \end{array}$

For example, find the equation of the tangent to the circle at point $\left(1,1\right)$ . The centre of the circle is at $\left(0,0\right)$ . The equation of the circle is ${x}^{2}+{y}^{2}=2$ .

Use

$y-{y}_{1}=-\frac{{x}_{1}-{x}_{0}}{{y}_{1}-{y}_{0}}\left(x-{x}_{1}\right)$

with $\left({x}_{0},{y}_{0}\right)=\left(0,0\right)$ and $\left({x}_{1},{y}_{1}\right)=\left(1,1\right)$ .

$\begin{array}{ccc}\hfill y-{y}_{1}& =& -\frac{{x}_{1}-{x}_{0}}{{y}_{1}-{y}_{0}}\left(x-{x}_{1}\right)\hfill \\ \hfill y-1& =& -\frac{1-0}{1-0}\left(x-1\right)\hfill \\ \hfill y-1& =& -\frac{1}{1}\left(x-1\right)\hfill \\ \hfill y& =& -\left(x-1\right)+1\hfill \\ \hfill y& =& -x+1+1\hfill \\ \hfill y& =& -x+2\hfill \end{array}$

## Co-ordinate geometry

1. Find the equation of the cicle:
1. with centre $\left(0;5\right)$ and radius 5
2. with centre $\left(2;0\right)$ and radius 4
3. with centre $\left(5;7\right)$ and radius 18
4. with centre $\left(-2;0\right)$ and radius 6
5. with centre $\left(-5;-3\right)$ and radius $\sqrt{3}$
1. Find the equation of the circle with centre $\left(2;1\right)$ which passes through $\left(4;1\right)$ .
2. Where does it cut the line $y=x+1$ ?
1. Find the equation of the circle with center $\left(-3;-2\right)$ which passes through $\left(1;-4\right)$ .
2. Find the equation of the circle with center $\left(3;1\right)$ which passes through $\left(2;5\right)$ .
3. Find the point where these two circles cut each other.
2. Find the center and radius of the following circles:
1. ${\left(x-9\right)}^{2}+{\left(y-6\right)}^{2}=36$
2. ${\left(x-2\right)}^{2}+{\left(y-9\right)}^{2}=1$
3. ${\left(x+5\right)}^{2}+{\left(y+7\right)}^{2}=12$
4. ${\left(x+4\right)}^{2}+{\left(y+4\right)}^{2}=23$
5. $3{\left(x-2\right)}^{2}+3{\left(y+3\right)}^{2}=12$
6. ${x}^{2}-3x+9={y}^{2}+5y+25=17$
3. Find the $x-$ and $y-$ intercepts of the following graphs and draw a sketch to illustrate your answer:
1. ${\left(x+7\right)}^{2}+{\left(y-2\right)}^{2}=8$
2. ${x}^{2}+{\left(y-6\right)}^{2}=100$
3. ${\left(x+4\right)}^{2}+{y}^{2}=16$
4. ${\left(x-5\right)}^{2}+{\left(y+1\right)}^{2}=25$
4. Find the center and radius of the following circles:
1. ${x}^{2}+6x+{y}^{2}-12y=-20$
2. ${x}^{2}+4x+{y}^{2}-8y=0$
3. ${x}^{2}+{y}^{2}+8y=7$
4. ${x}^{2}-6x+{y}^{2}=16$
5. ${x}^{2}-5x+{y}^{2}+3y=-\frac{3}{4}$
6. ${x}^{2}-6nx+{y}^{2}+10ny=9{n}^{2}$
5. Find the equations to the tangent to the circle:
1. ${x}^{2}+{y}^{2}=17$ at the point $\left(1;4\right)$
2. ${x}^{2}+{y}^{2}=25$ at the point $\left(3;4\right)$
3. ${\left(x+1\right)}^{2}+{\left(y-2\right)}^{2}=25$ at the point $\left(3;5\right)$
4. ${\left(x-2\right)}^{2}+{\left(y-1\right)}^{2}=13$ at the point $\left(5;3\right)$

## Rotation of a point about an angle $\theta$

First we will find a formula for the co-ordinates of $P$ after a rotation of $\theta$ .

We need to know something about polar co-ordinates and compound angles before we start.

## Polar co-ordinates

 Notice that : $sin\alpha =\frac{y}{r}$ $\therefore y=rsin\alpha$ and $cos\alpha =\frac{x}{r}$ $\therefore x=rcos\alpha$
so $P$ can be expressed in two ways:
1. $P\left(x;y\right)$ rectangular co-ordinates
2. $P\left(rcos\alpha ;rsin\alpha \right)$ polar co-ordinates.

## Compound angles

(See derivation of formulae in Ch. 12)

$\begin{array}{ccc}\hfill sin\left(\alpha +\beta \right)& =& sin\alpha cos\beta +sin\beta cos\alpha \hfill \\ \hfill cos\left(\alpha +\beta \right)& =& cos\alpha cos\beta -sin\alpha sin\beta \hfill \end{array}$

## Now consider ${P}^{\text{'}}$ After a rotation of $\theta$

$\begin{array}{ccc}& & P\left(x;y\right)=P\left(rcos\alpha ;rsin\alpha \right)\hfill \\ & & {P}^{\text{'}}\left(rcos\left(\alpha +\theta \right);rsin\left(\alpha +\theta \right)\right)\hfill \end{array}$
Expand the co-ordinates of ${P}^{\text{'}}$
$\begin{array}{ccc}\hfill x-\mathrm{co-ordinate of}{P}^{\text{'}}& =& rcos\left(\alpha +\theta \right)\hfill \\ & =& r\left[cos,\alpha ,cos,\theta ,-,sin,\alpha ,sin,\theta \right]\hfill \\ & =& rcos\alpha cos\theta -rsin\alpha sin\theta \hfill \\ & =& xcos\theta -ysin\theta \hfill \end{array}$
$\begin{array}{ccc}\hfill y-\mathrm{co-ordinate of P\text{'}}& =& rsin\left(\alpha +\theta \right)\hfill \\ & =& r\left[sin,\alpha ,cos,\theta ,+,sin,\theta ,cos,\alpha \right]\hfill \\ & =& rsin\alpha cos\theta +rcos\alpha sin\theta \hfill \\ & =& ycos\theta +xsin\theta \hfill \end{array}$

which gives the formula ${P}^{\text{'}}=\left[\left(,x,cos,\theta ,-,y,sin,\theta ,;,y,cos,\theta ,+,x,sin,\theta ,\right)\right]$ .

So to find the co-ordinates of $P\left(1;\sqrt{3}\right)$ after a rotation of 45 ${}^{\circ }$ , we arrive at:

$\begin{array}{ccc}\hfill {P}^{\text{'}}& =& \left[\left(,x,cos,\theta ,-,y,sin,\theta ,\right),;,\left(,y,cos,\theta ,+,x,sin,\theta ,\right)\right]\hfill \\ & =& \left[\left(1cos{45}^{\circ }-\sqrt{3}sin{45}^{\circ }\right),;,\left(\sqrt{3}cos{45}^{\circ }+1sin{45}^{\circ }\right)\right]\hfill \\ & =& \left[\left(\frac{1}{\sqrt{2}},-,\frac{\sqrt{3}}{\sqrt{2}}\right),;,\left(\frac{\sqrt{3}}{\sqrt{2}},+,\frac{1}{\sqrt{2}}\right)\right]\hfill \\ & =& \left(\frac{1-\sqrt{3}}{\sqrt{2}},;,\frac{\sqrt{3}+1}{\sqrt{2}}\right)\hfill \end{array}$

## Rotations

Any line $OP$ is drawn (not necessarily in the first quadrant), making an angle of $\theta$ degrees with the $x$ -axis. Using the co-ordinates of $P$ and the angle $\alpha$ , calculate the co-ordinates of ${P}^{\text{'}}$ , if the line $OP$ is rotated about the origin through $\alpha$ degrees.

 $P$ $\alpha$ 1. (2, 6) 60 ${}^{\circ }$ 2. (4, 2) 30 ${}^{\circ }$ 3. (5, -1) 45 ${}^{\circ }$ 4. (-3, 2) 120 ${}^{\circ }$ 5. (-4, -1) 225 ${}^{\circ }$ 6. (2, 5) -150 ${}^{\circ }$

## Characteristics of transformations

Rigid transformations like translations, reflections, rotations and glide reflections preserve shape and size, and that enlargement preserves shape but not size.

## Geometric transformations:

Draw a large 15 $×$ 15 grid and plot $▵ABC$ with $A\left(2;6\right)$ , $B\left(5;6\right)$ and $C\left(5;1\right)$ . Fill in the lines $y=x$ and $y=-x$ . Complete the table below , by drawing the images of $▵ABC$ under the given transformations. The first one has been done for you.

 Description Transformation (translation, reflection, Co-ordinates Lengths Angles rotation, enlargement) ${A}^{\text{'}}\left(2;-6\right)$ ${A}^{\text{'}}{B}^{\text{'}}=3$ ${\stackrel{^}{B}}^{\text{'}}={90}^{\circ }$ $\left(x;y\right)\to \left(x;-y\right)$ reflection about the $x$ -axis ${B}^{\text{'}}\left(5;-6\right)$ ${B}^{\text{'}}{C}^{\text{'}}=4$ $tan\stackrel{^}{A}=4/3$ ${C}^{\text{'}}\left(5;-2\right)$ ${A}^{\text{'}}{C}^{\text{'}}=5$ $\therefore \stackrel{^}{A}={53}^{\circ },\stackrel{^}{C}={37}^{\circ }$ $\left(x;y\right)\to \left(x+1;y-2\right)$ $\left(x;y\right)\to \left(-x;y\right)$ $\left(x;y\right)\to \left(-y;x\right)$ $\left(x;y\right)\to \left(-x;-y\right)$ $\left(x;y\right)\to \left(2x;2y\right)$ $\left(x;y\right)\to \left(y;x\right)$ $\left(x;y\right)\to \left(y;x+1\right)$

A transformation that leaves lengths and angles unchanged is called a rigid transformation. Which of the above transformations are rigid?

## Exercises

1. $\Delta ABC$ undergoes several transformations forming $\Delta {A}^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}$ . Describe the relationship between the angles and sides of $\Delta ABC$ and $\Delta {A}^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}$ (e.g., they are twice as large, the same, etc.)
 Transformation Sides Angles Area Reflect Reduce by a scale factor of 3 Rotate by 90 ${}^{\circ }$ Translate 4 units right Enlarge by a scale factor of 2
2. $\Delta DEF$ has $\stackrel{^}{E}={30}^{\circ }$ , $DE=4\phantom{\rule{0.166667em}{0ex}}\mathrm{cm}$ , $EF=5\phantom{\rule{0.166667em}{0ex}}\mathrm{cm}$ . $\Delta DEF$ is enlarged by a scale factor of 6 to form $\Delta {D}^{\text{'}}{E}^{\text{'}}{F}^{\text{'}}$ .
1. Solve $\Delta DEF$
2. Hence, solve $\Delta {D}^{\text{'}}{E}^{\text{'}}{F}^{\text{'}}$
3. $\Delta XYZ$ has an area of $6\phantom{\rule{0.166667em}{0ex}}{\mathrm{cm}}^{2}$ . Find the area of $\Delta {X}^{\text{'}}{Y}^{\text{'}}{Z}^{\text{'}}$ if the points have been transformed as follows:
1. $\left(x,y\right)\to \left(x+2;y+3\right)$
2. $\left(x,y\right)\to \left(y;x\right)$
3. $\left(x,y\right)\to \left(4x;y\right)$
4. $\left(x,y\right)\to \left(3x;y+2\right)$
5. $\left(x,y\right)\to \left(-x;-y\right)$
6. $\left(x,y\right)\to \left(x;-y+3\right)$
7. $\left(x,y\right)\to \left(4x;4y\right)$
8. $\left(x,y\right)\to \left(-3x;4y\right)$

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