Circles • Topic 2 of 3

Properties of Tangents & Perpendicularity Theorems

What is the Tangent Perpendicularity Theorem? The most important property of a tangent concerns its angle with the radius. Theorem (Tangent ⊥ Radius): The tangent at any point of a circle is perpendicular to the radius drawn to the point of contact. In symbols, if a tangent touches a circle (centre $O$) at $P$, then $OP\perp$ tangent, i.e. the angle between them is $90^{\circ}$.

Proof. Let the tangent touch the circle at $P$, and let $O$ be the centre. Take any other point $Q$ on the tangent (other than $P$). Since the tangent meets the circle only at $P$, every point of the tangent except $P$ lies outside the circle. Hence $Q$ is outside the circle, so $OQ>OP$ (a point inside the circle is closer to the centre than the radius; a point outside is farther). This is true for every choice of $Q$ on the line. Therefore $OP$ is the shortest distance from $O$ to the tangent. But the shortest distance from a point to a line is the perpendicular distance. Hence $OP\perp$ tangent, which proves $\angle OPQ = 90^{\circ}$. Q.E.D.

The Converse Theorem. The reverse is also true: if a line through a point $P$ on the circle is perpendicular to the radius $OP$, then that line is a tangent to the circle. Reason: since $OP$ is the shortest distance to the line, every other point of the line is farther than $r$ from $O$, hence outside the circle — so the line meets the circle only at $P$, which is the definition of a tangent.

Tangent length from an external point. Because $OP\perp PT$ (where $PT$ is a tangent touching at $T$ and $P$ is external), triangle $OTP$ is right-angled at $T$. By Pythagoras, $OP^2 = OT^2 + PT^2$, so the tangent length is

$PT = \sqrt{OP^2 - OT^2} = \sqrt{d^2 - r^2}$,

where $d=OP$ is the distance of the external point from the centre and $r$ is the radius. This single formula powers most numerical questions in this chapter.

Number of tangents from varying points. (1) From a point inside the circle: zero tangents, since every line through an interior point cuts the circle twice. (2) From a point on the circle: exactly one tangent (perpendicular to the radius at that point). (3) From a point outside the circle: exactly two tangents.

A useful chord result (two concentric circles). If a chord of a larger circle touches a smaller concentric circle, the point of contact is the midpoint of that chord, and the common-centre radius to the contact point is perpendicular to the chord. This lets you split the chord into two equal halves and apply Pythagoras — a favourite board exam set-up.

Why “perpendicular” matters numerically. Every standard distance computation in this chapter is just Pythagoras applied to the right angle at the point of contact. Memorise the three rearrangements of $OP^2 = r^2 + PT^2$: the tangent length $PT=\sqrt{OP^2-r^2}$; the radius $r=\sqrt{OP^2-PT^2}$; and the centre distance $OP=\sqrt{r^2+PT^2}$. Knowing which quantity is the hypotenuse ($OP$, always opposite the right angle at the contact point) prevents the single most common arithmetic error in board answers.

Tangents and symmetry. The converse theorem is what justifies the standard ruler-and-compass construction of a tangent at a given point: draw the radius to that point, then erect a perpendicular there. Because perpendicular-to-radius forces tangency, the construction is guaranteed correct without any further measurement — a clean example of a theorem doing practical work.

Common mistakes to avoid. (i) The perpendicular holds only at the point of contact, not at any random point of the tangent. (ii) When applying Pythagoras, $OP$ (centre to external point) is always the hypotenuse; mixing it up with the tangent length gives a wrong, larger answer. (iii) The distance “from the point to the circle” is not the same as $OP$; you must add the radius: $OP=(\text{distance to circle})+r$. (iv) Always confirm the right angle is at the point of contact before writing $a^2+b^2=c^2$.

Tangent to a circle is perpendicular to the radius at the point of contactTangent ⊥ radius at the point of contactOradiusP (point of contact)tangent90°Number of tangents from a point: zero from inside, one from on the circle, two from outsideHow many tangents from a point?PInside0 tangentsPOn the circle1 tangentPOutside2 tangents
13
Worked Example
A tangent $PQ$ touches a circle of radius $7$ cm (centre $O$) at $P$. If $OQ=25$ cm, find the tangent length $PQ$.
Solution
  1. Step 1: The radius $OP\perp PQ$, so $\triangle OPQ$ is right-angled at $P$ with hypotenuse $OQ$.
  2. Step 2: $OQ^2 = OP^2 + PQ^2 \Rightarrow 25^2 = 7^2 + PQ^2$.
  3. Step 3: $625 = 49 + PQ^2 \Rightarrow PQ^2 = 576$.
  4. Step 4: $PQ = \sqrt{576} = 24$ cm.

Answer: $PQ = 24$ cm.

14
Worked Example
From an external point $P$, a tangent touches a circle of radius $9$ cm at $T$. The distance from $P$ to the nearest point of the circle is $6$ cm. Find $PT$.
Solution
  1. Step 1: $OP = (\text{distance to circle}) + r = 6 + 9 = 15$ cm.
  2. Step 2: $OT\perp PT$, so $\triangle OTP$ is right-angled at $T$.
  3. Step 3: $OP^2 = OT^2 + PT^2 \Rightarrow 15^2 = 9^2 + PT^2$.
  4. Step 4: $225 = 81 + PT^2 \Rightarrow PT^2 = 144 \Rightarrow PT = 12$ cm.

Answer: $PT = 12$ cm.

15
Worked Example
A tangent line $XY$ touches a circle (centre $O$) at $T$. A chord $TA$ makes an angle of $40^{\circ}$ with the tangent direction $TX$. Find $\angle OTA$.
Solution
  1. Step 1: The radius $OT\perp XY$, so $\angle OTX = 90^{\circ}$.
  2. Step 2: $\angle OTX$ is split by $TA$ into $\angle OTA$ and $\angle ATX$.
  3. Step 3: Given $\angle ATX = 40^{\circ}$, we get $\angle OTA = 90^{\circ} - 40^{\circ}$.
  4. Step 4: $\angle OTA = 50^{\circ}$.

Answer: $\angle OTA = 50^{\circ}$.

16
Worked Example
State and prove the theorem: the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Solution
  1. Step 1: Let a tangent touch a circle with centre $O$ at point $P$. Take any other point $Q$ on the tangent.
  2. Step 2: Since the tangent meets the circle only at $P$, every other point of it lies outside the circle, so $OQ > OP$.
  3. Step 3: This holds for all $Q$ on the tangent, so $OP$ is the shortest distance from $O$ to the line.
  4. Step 4: The shortest distance from a point to a line is along the perpendicular, hence $OP\perp$ tangent, i.e. the angle is $90^{\circ}$.

Answer: Hence the tangent is perpendicular to the radius at the point of contact.

17
Worked Example
The length of the tangent from a point at distance $d$ from the centre is $\sqrt{d^2-r^2}$. Find the tangent length from a point $17$ cm from the centre of a circle of radius $8$ cm.
Solution
  1. Step 1: Tangent length $=\sqrt{d^2-r^2}$ with $d=17$, $r=8$.
  2. Step 2: $=\sqrt{17^2-8^2}=\sqrt{289-64}$.
  3. Step 3: $=\sqrt{225}$.
  4. Step 4: $=15$ cm.

Answer: $15$ cm.

18
Worked Example
A circle is drawn with $AB=12$ cm as diameter. A tangent at $B$ and a point $P$ on it satisfy $BP=5$ cm. Find $AP$, where $A$ and $B$ are the ends of the diameter.
Solution
  1. Step 1: $AB$ is a diameter, so $A$, $O$, $B$ are collinear and $AB\perp$ tangent at $B$ (the diameter through the contact point is along the radius).
  2. Step 2: Hence $\triangle ABP$ is right-angled at $B$.
  3. Step 3: $AP^2 = AB^2 + BP^2 = 12^2 + 5^2 = 144 + 25 = 169$.
  4. Step 4: $AP = \sqrt{169} = 13$ cm.

Answer: $AP = 13$ cm.

19
Worked Example
A line is drawn through a point $P$ on a circle (centre $O$) perpendicular to the radius $OP$. Prove that this line is a tangent.
Solution
  1. Step 1: Let the line through $P$ be perpendicular to $OP$. Take any other point $Q$ on this line.
  2. Step 2: In right $\triangle OPQ$ (right angle at $P$), $OQ$ is the hypotenuse, so $OQ > OP = r$.
  3. Step 3: Thus every point of the line other than $P$ is farther than $r$ from $O$, i.e. lies outside the circle.
  4. Step 4: The line therefore meets the circle only at $P$, which is the definition of a tangent.

Answer: Proved: the line is a tangent to the circle.

20
Worked Example
Two concentric circles have radii $13$ cm and $5$ cm. Find the length of the chord of the larger circle that touches the smaller circle.
Solution
  1. Step 1: The chord touches the inner circle, so the inner radius ($5$ cm) is perpendicular to the chord and bisects it.
  2. Step 2: With the outer radius $13$ cm as hypotenuse, half-chord $=\sqrt{13^2-5^2}$.
  3. Step 3: $=\sqrt{169-25}=\sqrt{144}=12$ cm.
  4. Step 4: Full chord $=2\times 12 = 24$ cm.

Answer: $24$ cm.

21
Worked Example
A tangent $PT$ of length $24$ cm is drawn from a point $P$ to a circle. If $OP=25$ cm, find the radius of the circle.
Solution
  1. Step 1: $OP^2 = r^2 + PT^2$ with $OP=25$, $PT=24$.
  2. Step 2: $25^2 = r^2 + 24^2 \Rightarrow 625 = r^2 + 576$.
  3. Step 3: $r^2 = 625-576 = 49$.
  4. Step 4: $r = \sqrt{49} = 7$ cm.

Answer: Radius $=7$ cm.

22
Worked Example
In the figure, $PT$ is a tangent and $PAB$ is a secant through the centre $O$ of a circle of radius $6$ cm. If $PA=4$ cm (with $A$ the nearer intersection), find $PT$.
Solution
  1. Step 1: The secant passes through the centre, so $AB$ is a diameter $=2r=12$ cm and $PB=PA+AB=4+12=16$ cm.
  2. Step 2: $OP = PA + OA = 4 + 6 = 10$ cm.
  3. Step 3: Tangent length $PT=\sqrt{OP^2-r^2}=\sqrt{10^2-6^2}=\sqrt{100-36}=\sqrt{64}$.
  4. Step 4: $PT = 8$ cm.

Answer: $PT = 8$ cm.

23
Worked Example
A tangent at point $A$ of a circle (centre $O$) meets a line through $O$ at $B$ such that $\angle ABO=30^{\circ}$ and the radius $OA=5$ cm. Find $OB$.
Solution
  1. Step 1: $OA\perp AB$ (tangent $\perp$ radius), so $\triangle OAB$ is right-angled at $A$.
  2. Step 2: $\sin(\angle ABO) = \dfrac{OA}{OB}$, i.e. $\sin 30^{\circ} = \dfrac{5}{OB}$.
  3. Step 3: $\dfrac{1}{2} = \dfrac{5}{OB} \Rightarrow OB = 10$ cm.
  4. Step 4: State the result.

Answer: $OB = 10$ cm.

24
Worked Example
A point $Q$ is at distance $5$ cm from the centre of a circle of radius $4$ cm. Find the length of the tangent from $Q$, and explain why $Q$ must be an external point.
Solution
  1. Step 1: Since $OQ=5>4=r$, the point $Q$ lies outside the circle, so a real tangent exists.
  2. Step 2: Tangent length $=\sqrt{OQ^2-r^2}=\sqrt{5^2-4^2}$.
  3. Step 3: $=\sqrt{25-16}=\sqrt{9}=3$ cm.
  4. Step 4: A tangent length is real only when $d>r$; here $5>4$, confirming $Q$ is external.

Answer: Tangent length $=3$ cm; $Q$ is external because $OQ=5>r=4$.

Key Points

  • The radius connecting to a tangent line at its point of contact always forms a 90-degree perpendicular angle.
  • This perpendicular relationship allows you to solve for unknown lengths using the Pythagoras Theorem.
  • Zero tangents can ever be drawn originating from an interior point located inside a circle.
  • Exactly one tangent line can be constructed passing through a point resting on the boundary.
  • Exactly two tangents can be projected onto a circle from any single exterior point.
  • Tangent length from an external point $=\sqrt{d^2-r^2}$, where $d$ is the centre-to-point distance and $r$ the radius.
  • Converse: a line through a point on the circle, perpendicular to the radius there, must be a tangent.
Tap an option to check your answer0 / 4
Q1.The tangent at any point of a circle is ___ to the radius through the point of contact.
Explanation: Tangent $\perp$ radius.
Q2.Tangents drawn from an external point to a circle are:
Explanation: $PA=PB$.
Q3.The angle between a tangent and the radius at the point of contact is:
Explanation: $90^\circ$.
Q4.If $PA$ and $PB$ are tangents from $P$, then:
Explanation: Equal tangent lengths.