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Points on the circle centre C and radius r are given by the equation $|Z-C|=r$ or $(Z-C)(\overline{Z}-\overline{C})=r^2$. Where $Z = x + iy$.

When multiplied out, I understand that we have

$$Z\overline{Z}-C\overline{Z}-\overline{C}Z+C\overline{C}=r^2\;.$$

So the question I am stuck on states that

If $B$ lies on the circle centre $C$ and radius $r$, show that the equation of the tangent at $B$ is:

$$(\overline{B}-\overline{C})Z+(B-C)\overline{Z}={B}\overline{B}- {C}\overline{C}+r^2\;.$$

Can anyone give me pointers?

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2 Answers 2

up vote 1 down vote accepted

First step

Show that $Z=B$ solves the equation, so $B$ lies on the assumed tangent line.

$$ \bigl(\bar B-\bar C\bigr)B + \bigl(B-C\bigr)\bar B = B\bar B - C\bar C+r^2 \\ \bar BB-\bar CB + B\bar B-C\bar B = B\bar B - C\bar C+r^2 \\ \bar CB + B\bar B-C\bar B + C\bar C = r^2 $$

So you arrive at the equation for the circle. If $B$ lies on that circle, then it will also lie on the line.

Second step

Show that the only $Z$ which satisfies both the line equation and the circle equation is $B$ itself. Assume $Z$ to lie on the circle, then plug it into the equation of the line.

\begin{gather*} \bigl(\bar B-\bar C\bigr)Z + \bigl(B-C\bigr)\bar Z = B\bar B - C\bar C+r^2 \\ \bar BZ-\bar CZ + B\bar Z-C\bar Z = B\bar B - C\bar C+r^2 \\ \bigl(\bar BZ+B\bar Z-B\bar B-Z\bar Z\bigr)+\bigl(Z\bar Z-\bar CZ - C\bar Z+C\bar C\bigr) = r^2 \tag{$*$} \\ \bar BZ+B\bar Z-B\bar B-Z\bar Z = 0 \\ B = Z \end{gather*}

Equation $(*)$ basically states that if $Z$ lies on the given circle around $C$, then $B$ lies on a circle of radius $0$ around $Z$, so the two must be identical. Therefore $Z=B$ is the only intersection of line and circle, and a line which intersects a circle in just one point is a tangent.

Alternative second step

You can even interpret $(*)$ as

$$\lvert Z-C\rvert^2 - \lvert Z-B\rvert^2 = r^2 = \lvert B-C\rvert^2$$

which you can see as the Pythagorean theorem, with $ZC$ the hypothenuse and $ZB$ and $BC$ the legs. So you can tell that $ZB$ is orthogonal to $BC$, which is another characterization of a tangent.

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this was extremely helpful, thanks!!! –  Jay C Oct 21 '13 at 6:24
    
@Jay: Please remember to accept the answer which was most useful to you. You can always change your mind if a better answer comes along. –  MvG Oct 21 '13 at 13:35

Suppose that $C=0$, that $r=1$ and that $B=1$. The equation is the $Z+\bar Z=2$ or, equivalently, $2\operatorname{Re}(Z)=2$. This is obviously the equation of the tangent to the unit circle at $1$, as we want.

Now to go from there to the general case, use a rotation, a homothecy and a translation.

Later. I don't know why you think this is not concrete :-)

Suppose $\mathcal C$ is the circle of radius $r$ centered at $C$ and $B$ is a point in $\mathcal C$.

  • The map $t:z\in\mathbb C\mapsto z\in \mathbb C$ is a translation. Let $\mathcal C'=t(\mathcal C)$, $C'=t(C)$ and $B'=t(B)$. Then it is easy to check that a point $Z$ belongs to the tangent to $\mathcal C$ at $B$ if and only iff the point $t(Z)$ belongs to the tangent to $\mathcal C'$ at $B'$. Notice that $\mathcal C'$ is a circle of radius $r$ centered at the origin, and that $B'=B-C$.

  • The map $h:z\in\mathbb C\mapsto z/r\in\mathbb C$ is a homothecy. Let $\mathcal C''=h(\mathcal C')$, $C''=h(C')$ and $B''=h(B')$. Then it is easy to check that a point $Z'$ belongs to the tangent to $\mathcal C'$ at $B'$ if and only iff the point $h(Z')$ belongs to the tangent to $\mathcal C''$ at $B''$. Notice that $\mathcal C''$ is a circle of radius $1$ centered at the origin, and that $B''$, being one of its points, has modulus $1$. in fact, $B''=(B-C)/r$.

  • The map $r:z\in\mathbb C\mapsto z\overline{B''}\in\mathbb C$ is a rotation around the origin (because $B''$ has modulus $1$). Let $\mathcal C'''=r(\mathcal C'')$, $C'''=r(C'')$ and $B'''=r(B'')$. Then it is easy to check that a point $Z''$ belongs to the tangent to $\mathcal C''$ at $B''$ if and only iff the point $r(Z'')$ belongs to the tangent to $\mathcal C'''$ at $B'''$. Notice that $\mathcal C'''$ is just $\mathcal C''$, the unit circle centered at the origin, and that $B'''$ is the point $1$.

In all, we have concluded that a point $Z$ belongs to the tangent to $\mathcal C$ at $B$ iff $r(h(t(Z)))=(Z-C)(\bar B-\bar C)/r^2$ belongs to the tangent to the unit circle at $1$. As we say before, this happens exactly if the real part of $(Z-C)(\bar B-\bar C)/r^2$ is $2$, that is, if $$(Z-C)(\bar B-\bar C)/r^2+(\bar Z-\bar C)(B-C)/r^2=2.$$ Moving the $r^2$ to the right hand side and expanding a bit that is left on the left,, we get the equivalent equation $$Z(\bar B-\bar C)+\bar Z(B-C)=2r^2+C(\bar B-\bar C)+\bar C(B-C). \tag{1}$$ Since $B$ is in the circle $\mathcal C$, we have $r^2=(B-C)(\bar B-\bar C)$, and using this to massage a bit the right hand side of (1) gives you the equation you want.

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Haha, I guess I didn't understand completely! I think this answer is definitely over my level/what I've learned though in class :) –  Jay C Oct 21 '13 at 9:55
    
You have learned everything in this answer, trust me. –  Mariano Suárez-Alvarez Oct 21 '13 at 15:09

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