# Construction of Tangent from Point to Circle

## Theorem

From a given point outside a given circle, it is possible to draw a tangent to that circle.

In the words of Euclid:

From a given point to draw a straight line touching a given circle.

## Proof 1 Let $A$ be the given point and let $BCD$ be the given circle.

It is required that a straight line be drawn from $A$ to $BCD$.

Join $AE$ and draw the circle $AFG$ with center $E$ and radius $AE$.

Join $EF$ and let $B$ be the point at which $EF$ joins the circle $BCD$.

Join $AB$.

Then $AB$ is the required tangent to $BCD$.

### Proof of Construction

Since $E$ is the center of the circles $BCD$ and $AFG$, $EA = EF$ and $ED = EB$.

Therefore the two sides $AE, EB$ equal the two sides $FE, ED$, and they contain a common angle at $E$.

So by Triangle Side-Angle-Side Equality, $\triangle DEF = \triangle BEA$.

Therefore $\angle EDF = \angle EBA$.

But $\angle EDF$ is a right angle, so $\angle EBA$ is also a right angle.

But $EB$ is a radius of circle $BCD$.

It follows from the porism to Line at Right Angles to Diameter of Circle that $AB$ is tangent to $BCD$.

$\blacksquare$

## Proof 2 Let $BCD$ with center $A$ be the circle, and let $E$ be the exterior point from which a tangent is to be drawn.

Bisect $AE$ at $F$.

Then draw a circle $AEG$ whose center is $F$ and whose radius is $AF$.

The point $G$ is where $AEG$ intersects $BCD$.

The line $EG$ is the required tangent.

### Proof of Construction

By the method of construction, $AE$ is the diameter of $AEG$.

By Thales' Theorem $\angle AGE$ is a right angle.

But $AG$ is a radius of $BCD$.

The result follows from Radius at Right Angle to Tangent.

$\blacksquare$