Circle

 Circle

Here’s a proper circle with its radius and tangent drawn.

• $\mathrm{<br>}$ is the centre,

• $\mathrm{<br>}$ is the point of contact,

• The green line is the radius $\mathrm{<br>}\mathrm{<i\; style="font-weight:\; bold;"><span\; class="katex"><span\; aria-hidden="true"\; class="katex-html"><span\; class="base"><span\; class="mord\; mathnormal">OP</span></span></span></span>,</i>}$

• The red line is the tangent at $\mathrm{<br>}$.

Here’s the circle with centre $\mathrm{<br>}$ and an external point $\mathrm{<br>}$.

• The two red lines are the tangents $\mathrm{<br>}$ and $\mathrm{<br>}$.

• The dashed green lines show the radii $OA$ and $\mathrm{ }$, each perpendicular to the tangents at the points of contact $A$ and  $B$.

• Statement.

• Let $\mathrm{<br>}$ be the centre of a circle and  $P$ an external point. Tangents $\mathrm{<br>}$ and $PB$ touch the circle at $\mathrm{ }$ and  $B$ respectively.Then $\mathrm{ }$.

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  Proof (using the labelled diagram)

  1. From the diagram, $\mathrm{ }$ and $\mathrm{ }$ are radii and meet the circle at  $A$ and $\mathrm{ }$.

  2. A radius to the point of contact is perpendicular to the tangent, so

     $$OA \perp PA\quad\text{and}\quad OB \perp PB.$$ 
     

     Thus  $△OAP$ and  $△OBP$ are right triangles (right angles at  $A$ and  $B$).

  3. In triangles  $△OAP$ and $\triangle OBP$:

     •  OA=OB (radii of the same circle),

     • $OP = OP$ (common side),

     • $\angle OAP = \angle OBP = 90^\circ$.

  4. By the RHS (Right angle–Hypotenuse–Side) congruence criterion,

     $$\triangle OAP \cong \triangle OBP.$$

  5. Corresponding parts of congruent triangles are equal, so

     $$PA = PB.$$

  ∴ Tangents drawn from an external point to a circle are equal in length.  

• Corollaries of “Tangents from an External Point are Equal”

  1. Equal Tangents

     • From an external point $P$, the two tangents $PA$ and $PB$ to a circle are equal in length.

     $$PA = PB$$ 

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  2. Tangents are Symmetrical

     • The line joining the centre  $O$ to the external point $P$ bisects the angle between the two tangents.

     $$\angle OAP = \angle OBP$$ 

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  3. Equal Inclination

     • The two tangents are equally inclined to the line $OP$ (joining centre to external point).

     $$\angle PAO = \angle PBO$$ 

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  4. Quadrilateral is Cyclic

     • The quadrilateral $OAPB$ is a cyclic quadrilateral because

     $$\angle OAP + \angle OBP = 180^\circ$$ 

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  5. Perpendicular from Centre Bisects the Angle

     • The line  $OP$ (joining centre and external point) bisects  $\angle APB$.

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  6. Tangent Length Formula

     • If $d = OP$ (distance of external point from centre) and $r = OA$ (radius), then

     $$$$PA=PB=d