The potency of a point in relation to a circle

Case 1: Internal point of a circle

If a point $ O$ is the intersection between two ropes $ (AB )$ and $ (CD )$ of a circle inside it, and such as the following figure:

Potency of a point in relation to a circle - internal point

So, the two triangles $ AOC $ and $ DOB $ are similar, and the product between the lengths respectively proportional starting from the point $O$ is worth:

$$\overline{OA} \times \hspace{0.05em} \overline{OB} =\hspace{0.05em} \overline{OC} \times \overline{OD}$$

Case 2: External point of a circle

If a point $ O$ is the intersection between two ropes $ (AC )$ and $ (DB )$ of a circle outside it, and such as the following figure:

Potency of a point in relation to a circle - external point

So, the two triangles $ ABO $ and $ DCO $ are similar, and the product between the lengths respectively proportional starting from the point $O$ is worth:

$$\overline{OA} \times \hspace{0.05em} \overline{OB} =\hspace{0.05em} \overline{OC} \times \overline{OD}$$

Case 3: External point of a circle with one tangent

If a point $ O$ is the intersection between a rope $ (AC )$ of a circle and a tangent to the same circle passing through $ T$, and such as the following figure:

Potency of a point in relation to a circle - external point and one tangent rope

So, the two triangles $ OAT $ and $ CTO $ are similar, and the product between the lengths respectively proportional starting from the point $O$ is worth:

$$\overline{OA} \times \hspace{0.05em} \overline{OC} =\hspace{0.05em} \overline{OT} ^2$$

Case 4: External point of a circle with two tangents

If a point $ O$ is the intersection between two tangents to the circle passing respectively through $ T$ and $ T'$, and such as the following figure:

Potency of a point in relation to a circle - external point and two tangent ropes

So, the triangle $ OTT' $ is an isosceles triangle, and in this case:

$$\overline{OT} = \hspace{0.05em} \overline{OT'}$$

Demonstrations

Case 1: Internal point of a circle

Let us consider a circle and two ropes $AB$ and $CD$ such as the following figure:

Potency of a point in relation to a circle - internal point - demo 1

Point $O$ is the point of intersection between the lines $(AB)$ and $(CD)$ inside the circle.

Angles $ \widehat{BAC}$ and $ \widehat{BDC}$ intercept the same arc $ \overset{\frown}{BC}$ on the circle. They are therefore equal.

$$ \widehat{BAC} = \widehat{BDC} = \alpha $$

In the same way, the angles $ \widehat{ABD}$ and $ \widehat{ACD}$ intercept the same arc $ \overset{\frown}{AD}$ on the circle.

$$ \widehat{ABD} = \widehat{ACD} = \beta $$

Potency of a point in relation to a circle - internal point - demo 2

The two triangles $ AOC $ and $ DOB $ having two angles two by two respectively equal, they are similar.

We then have the following proportions:

$$ \frac{OA}{OC} = \frac{OD}{OB} $$

And as a result,

$$\overline{OA} \times \hspace{0.05em} \overline{OB} =\hspace{0.05em} \overline{OC} \times \overline{OD}$$

Case 2: External point of a circle

Let us consider a circle and two ropes $AB$ and $CD$ such as the following figure:

Potency of a point in relation to a circle - external point - demo 1

Point $O$ is the point of intersection between the lines $(AB)$ and $(CD)$ outside the circle, this intersection forms an angle vertex $\theta$.

Angles $ \widehat{BAC}$ and $ \widehat{BDC}$ intercept the same arc $ \overset{\frown}{BC}$ on the circle. They are therefore equal.

$$ \widehat{BAC} = \widehat{BDC} = \alpha $$

Consequently, the angle measures $ \widehat{ABO}$ and $ \widehat{DCO}$ are equals, being the third angle of the two respective triangles:

$$ \widehat{ABO} = \widehat{DCO} = \beta = \pi - (\alpha + \theta) $$

Potency of a point in relation to a circle - external point - demo 2

The two triangles $ ABO $ and $ DCO $ having two angles two by two respectively equal, they are similar.

We then have the following proportions:

$$ \frac{OA}{OC} = \frac{OD}{OB} $$

And as a result,

$$\overline{OA} \times \hspace{0.05em} \overline{OB} =\hspace{0.05em} \overline{OC} \times \overline{OD}$$

Case 3: External point of a circle with one tangent

By returning to our previous figure by making tend the points $B$ and $D$ towards a new point $T$, so that the straight line $(OT)$ be a tangent to the circle.

Potency of a point in relation to a circle - external point and one tangent rope - demo 1

By constructing this new point $T$, we retained the equivalence between the two angles $\alpha$.

Potency of a point in relation to a circle - external point and one tangent rope - demo 2

The two triangles $ OAT $ an $ CTO $ having two angles two by two respectively equal, they are similar.

We then have the following proportions:

$$ \frac{OA}{OT} = \frac{OT}{OC} $$

And as a result,

$$\overline{OA} \times \hspace{0.05em} \overline{OC} =\hspace{0.05em} \overline{OT} ^2$$

Case 4: External point of a circle with two tangents

By returning to our previous figure by making tend the points this time the points $A$ and $C$ towards a new point $T'$, so that the straight line $(OT')$ be a tangent to the circle.

Potency of a point in relation to a circle - external point and two tangent ropes - demo 1

By constructing this new point $T'$, we retained the equivalence between the two angles $\alpha$.

Potency of a point in relation to a circle - external point and two tangent ropes - demo 2

The two similar triangles having merged during sliding, we then find ourselves with a single isosceles triangle $OTT'$.

And as a result,

$$\overline{OT} = \hspace{0.05em} \overline{OT'}$$