What is the Length-of-Tangents Theorem? From a point outside a circle you can draw exactly two tangents. Theorem: The lengths of the two tangents drawn from an external point to a circle are equal.
Proof. Let $PA$ and $PB$ be the two tangents from external point $P$ to a circle with centre $O$, touching at $A$ and $B$. Join $OA$, $OB$ and $OP$. In triangles $OAP$ and $OBP$:
- $OA = OB$ (radii of the same circle),
- $\angle OAP = \angle OBP = 90^{\circ}$ (tangent $\perp$ radius at the point of contact),
- $OP = OP$ (common hypotenuse).
Hence $\triangle OAP \cong \triangle OBP$ by the RHS congruence rule. By CPCT, $PA = PB$, so the two tangent lengths are equal. Q.E.D. Two further consequences follow at once from the same congruence: $\angle APO = \angle BPO$ (so $OP$ bisects the angle between the tangents) and $\angle AOP = \angle BOP$ (so $OP$ bisects the angle between the radii).
Angle between tangents and the central angle. Consider the quadrilateral $OAPB$. Its four angles add to $360^{\circ}$. Two of them are right angles at $A$ and $B$, so
$\angle AOB + \angle APB + 90^{\circ} + 90^{\circ} = 360^{\circ} \;\Rightarrow\; \angle AOB + \angle APB = 180^{\circ}.$
That is, the angle between the two tangents and the angle subtended at the centre by the radii are supplementary.
Triangle & quadrilateral applications. Because two tangents from a vertex are equal, problems about circles inscribed in polygons reduce to simple algebra. For a triangle $ABC$ with an inscribed circle (incircle) touching $BC,CA,AB$ at $D,E,F$: $AE=AF$, $BD=BF$, $CD=CE$. Adding around the triangle gives each tangent length as $s-a$, $s-b$, $s-c$ where $s$ is the semi-perimeter.
Tangential quadrilateral. If a quadrilateral $ABCD$ circumscribes a circle, then the sums of its pairs of opposite sides are equal: $AB+CD = BC+DA$. Proof: label the tangent lengths from $A,B,C,D$ as $p,q,r,s$. Then $AB=p+q$, $BC=q+r$, $CD=r+s$, $DA=s+p$, so $AB+CD=p+q+r+s=BC+DA$. A neat corollary: a parallelogram that circumscribes a circle must be a rhombus (opposite sides equal forces all four sides equal).
For a right triangle right-angled with legs $a,b$ and hypotenuse $c$, the inradius is $r=\dfrac{a+b-c}{2}$ — this drops straight out of the equal-tangent relations and is a popular olympiad shortcut.
Incircle tangent lengths and the semi-perimeter. For a triangle with sides $a=BC$, $b=CA$, $c=AB$ and semi-perimeter $s=\dfrac{a+b+c}{2}$, the tangent lengths from the vertices are $s-a$ (from $A$), $s-b$ (from $B$) and $s-c$ (from $C$). This compact result lets you read off every tangent length the moment you know the three sides, and it underpins the area formula $\text{Area}=rs$ linking inradius $r$, semi-perimeter $s$ and area.
The third-tangent (perimeter) trick. When a tangent cuts across the two tangents from $P$, the new contact point creates two more equal-tangent pairs. The perimeter of the small triangle so formed always collapses to $PA+PB=2\,PA$ — recognising this saves a great deal of algebra in exam questions and olympiad problems alike.
Common mistakes to avoid. (i) Equal-tangent pairs are measured from the same vertex/point — do not equate tangents from different vertices. (ii) In a tangential quadrilateral the equal sums are of opposite sides, not adjacent ones. (iii) $\angle AOB$ and $\angle APB$ are supplementary, not equal — a very common slip. (iv) When a third tangent cuts across two others (as in perimeter problems), remember the new contact point again creates two equal sub-segments.
From an external point $P$, tangents $PA$ and $PB$ are drawn to a circle. If $PA=14$ cm, find $PB$.
Solution- Step 1: Tangents drawn from the same external point to a circle are equal in length.
- Step 2: $PA$ and $PB$ are both from $P$, so $PB=PA$.
- Step 3: Therefore $PB=14$ cm.
Answer: $PB=14$ cm.
Tangents $PA$ and $PB$ are drawn from $P$ to a circle with centre $O$. If $\angle APB=70^{\circ}$, find the central angle $\angle AOB$.
Solution- Step 1: In quadrilateral $OAPB$, $\angle OAP=\angle OBP=90^{\circ}$ (radius $\perp$ tangent).
- Step 2: Angle sum: $\angle AOB + \angle APB + 90^{\circ}+90^{\circ}=360^{\circ}$.
- Step 3: $\angle AOB + \angle APB = 180^{\circ}$, so $\angle AOB = 180^{\circ}-70^{\circ}$.
- Step 4: $\angle AOB = 110^{\circ}$.
Answer: $\angle AOB = 110^{\circ}$.
A circle is inscribed in $\triangle ABC$, touching $AB$, $BC$, $CA$ at $D$, $E$, $F$. If $AD=5$ cm, $BE=6$ cm and $CF=7$ cm, find the perimeter of $\triangle ABC$.
Solution- Step 1: Equal tangents from each vertex: $AD=AF=5$, $BD=BE=6$, $CE=CF=7$ (all in cm).
- Step 2: $AB=AD+BD=5+6=11$ cm.
- Step 3: $BC=BE+CE=6+7=13$ cm; $CA=CF+AF=7+5=12$ cm.
- Step 4: Perimeter $=11+13+12=36$ cm.
Answer: Perimeter $=36$ cm.
Tangents $PA$ and $PB$ are drawn from $P$ to a circle of centre $O$. Prove that $\angle PTQ$-type result: prove $PA=PB$ using congruent triangles.
Solution- Step 1: Join $OA$, $OB$, $OP$. Then $OA=OB$ (radii) and $\angle OAP=\angle OBP=90^{\circ}$ (radius $\perp$ tangent).
- Step 2: $OP=OP$ is common.
- Step 3: By RHS congruence, $\triangle OAP\cong\triangle OBP$.
- Step 4: By CPCT, $PA=PB$.
Answer: Proved: $PA=PB$ (tangents from an external point are equal).
A quadrilateral $ABCD$ circumscribes a circle. Prove that $AB+CD=BC+DA$.
Solution- Step 1: Let the circle touch $AB,BC,CD,DA$ at $P,Q,R,S$. From each vertex the two tangents are equal: $AP=AS$, $BP=BQ$, $CQ=CR$, $DR=DS$.
- Step 2: $AB+CD=(AP+BP)+(CR+DR)$ and $BC+DA=(BQ+CQ)+(DS+AS)$.
- Step 3: Substitute equal tangents: $AB+CD=AP+BP+CR+DR$ and $BC+DA=BP+CR+DR+AP$ (same four terms).
- Step 4: Hence $AB+CD=BC+DA$.
Answer: Proved: $AB+CD=BC+DA$.
A circle is inscribed in $\triangle ABC$ with $AB=12$ cm, $BC=8$ cm, $CA=10$ cm. Find the tangent lengths from $A$, $B$ and $C$.
Solution- Step 1: Semi-perimeter $s=\dfrac{12+8+10}{2}=15$ cm.
- Step 2: Tangent from $A$ $=s-a$, where $a=BC=8$: $15-8=7$ cm.
- Step 3: Tangent from $B$ $=s-b$, where $b=CA=10$: $15-10=5$ cm.
- Step 4: Tangent from $C$ $=s-c$, where $c=AB=12$: $15-12=3$ cm.
Answer: From $A$: $7$ cm; from $B$: $5$ cm; from $C$: $3$ cm.
Prove that a parallelogram circumscribing a circle is a rhombus.
Solution- Step 1: Let parallelogram $ABCD$ circumscribe a circle. As a tangential quadrilateral, $AB+CD=BC+DA$.
- Step 2: In a parallelogram opposite sides are equal: $AB=CD$ and $BC=DA$.
- Step 3: Substituting, $2AB=2BC$, so $AB=BC$.
- Step 4: All four sides are then equal, so $ABCD$ is a rhombus.
Answer: Proved: the parallelogram is a rhombus.
Two tangents $PA$ and $PB$ are drawn from $P$ to a circle of centre $O$. A third tangent touches the circle at $C$ and meets $PA$ at $D$ and $PB$ at $E$. If $PA=10$ cm, find the perimeter of $\triangle PDE$.
Solution- Step 1: From external points, $DA=DC$ and $EC=EB$ (equal tangents).
- Step 2: Perimeter of $\triangle PDE = PD+DE+EP = PD+(DC+CE)+EP$.
- Step 3: Replace: $DC=DA$ and $CE=EB$, so it becomes $(PD+DA)+(EP+EB)=PA+PB$.
- Step 4: Since $PA=PB=10$ cm, perimeter $=10+10=20$ cm.
Answer: Perimeter of $\triangle PDE = 20$ cm.
A right triangle has legs $6$ cm and $8$ cm. Find the radius of its inscribed circle.
Solution- Step 1: Hypotenuse $=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$ cm.
- Step 2: For a right triangle the inradius $r=\dfrac{a+b-c}{2}$ with legs $a,b$ and hypotenuse $c$.
- Step 3: $r=\dfrac{6+8-10}{2}=\dfrac{4}{2}$.
- Step 4: $r=2$ cm.
Answer: Inradius $=2$ cm.
A circle touches the four sides of quadrilateral $ABCD$ with $AB=x+3$, $BC=2x$, $CD=x+5$, $DA=x+1$ (in cm). Find $x$ and the side $BC$.
Solution- Step 1: For a tangential quadrilateral, $AB+CD=BC+DA$.
- Step 2: $(x+3)+(x+5)=2x+(x+1)$, i.e. $2x+8=3x+1$.
- Step 3: $8-1=3x-2x \Rightarrow x=7$.
- Step 4: $BC=2x=14$ cm.
Answer: $x=7$; $BC=14$ cm.
Tangents $PA$ and $PB$ from $P$ to a circle (centre $O$) make $\angle APB=120^{\circ}$. Find the ratio $OP:PA$.
Solution- Step 1: $OP$ bisects $\angle APB$, so $\angle APO=60^{\circ}$, and $\angle OAP=90^{\circ}$.
- Step 2: In right $\triangle OAP$, $\cos(\angle APO)=\dfrac{PA}{OP}$.
- Step 3: $\cos 60^{\circ}=\dfrac{1}{2}=\dfrac{PA}{OP}$, so $OP=2\,PA$.
- Step 4: Hence $OP:PA=2:1$.
Answer: $OP:PA = 2:1$.
Two tangents $TP$ and $TQ$ are drawn from an external point $T$ to a circle with centre $O$. Prove that $\angle PTQ = 2\,\angle OPQ$.
Solution- Step 1: Let $\angle PTQ=\theta$. Since $TP=TQ$ (equal tangents), $\triangle TPQ$ is isosceles, so $\angle TPQ=\angle TQP=\dfrac{180^{\circ}-\theta}{2}=90^{\circ}-\dfrac{\theta}{2}$.
- Step 2: $OP\perp TP$ (radius $\perp$ tangent), so $\angle OPT=90^{\circ}$.
- Step 3: $\angle OPQ=\angle OPT-\angle TPQ=90^{\circ}-\left(90^{\circ}-\dfrac{\theta}{2}\right)=\dfrac{\theta}{2}$.
- Step 4: Therefore $\angle PTQ=\theta=2\times\dfrac{\theta}{2}=2\,\angle OPQ$.
Answer: Proved: $\angle PTQ = 2\,\angle OPQ$.