Circles — Important Questions

Answer: (a) $0$

Every line through an interior point is a secant (it cuts the circle in two points), so no tangent can be drawn from a point inside a circle.

Answer: (a) perpendicular to the radius through the point of contact.

This is the fundamental tangent theorem: the radius drawn to the point of contact is perpendicular to the tangent there.

Answer: (a) $12$ cm

The radius $OP\perp PQ$, so triangle $OPQ$ is right-angled at $P$. Hence $PQ=\sqrt{OQ^2-OP^2}=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12$ cm.

Answer: (a) $8$ cm

The chord touches the inner circle, so the perpendicular from the centre (length $=3$ cm, the inner radius) bisects the chord. Half the chord $=\sqrt{5^2-3^2}=\sqrt{16}=4$ cm, so the chord $=2\times4=8$ cm.

Answer: (b) Both A and R are true but R is not the correct explanation of A.

Both statements are true theorems. The equal-tangent result (A) is proved using the perpendicularity (R) together with the common hypotenuse $OP$ and equal radii; however R by itself is a separate property and is not, on its own, the complete reason A holds. So R does not fully explain A.

In quadrilateral $OAPB$, the radii meet the tangents at right angles, so $\angle OAP=\angle OBP=90^\circ$.

The angle sum of a quadrilateral is $360^\circ$:

$\angle AOB+\angle APB+\angle OAP+\angle OBP=360^\circ$

$\angle AOB+60^\circ+90^\circ+90^\circ=360^\circ\Rightarrow \angle AOB=120^\circ$.

Let the tangents touch at $A$ and $B$. $OP$ bisects $\angle APB$, so $\angle APO=30^\circ$, and $OA\perp PA$ so triangle $OAP$ is right-angled at $A$ with $OA=3$ cm.

$\tan(\angle APO)=\dfrac{OA}{PA}\Rightarrow \tan30^\circ=\dfrac{3}{PA}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{3}{PA}$.

Therefore $PA=3\sqrt3$ cm. Each tangent is $3\sqrt3\approx5.19$ cm long.

Given: A circle with centre $O$ and an external point $P$; $PA$ and $PB$ are tangents touching the circle at $A$ and $B$.

To prove: $PA=PB$.

Construction: Join $OA$, $OB$ and $OP$.

Proof: Since a tangent is perpendicular to the radius at the point of contact, $\angle OAP=\angle OBP=90^\circ$.

In right triangles $OAP$ and $OBP$:

• $OA=OB$ (radii of the same circle),
• $OP=OP$ (common hypotenuse),
• $\angle OAP=\angle OBP=90^\circ$.

By the RHS congruence rule, $\triangle OAP\cong\triangle OBP$.

Therefore, by CPCT, $PA=PB$. Hence the tangents from an external point are equal in length.

Let the circle touch $AB,BC,CD,DA$ at $P,Q,R,S$ respectively.

Tangents drawn from an external point are equal, so:

• From $A$: $AP=AS$
• From $B$: $BP=BQ$
• From $C$: $CR=CQ$
• From $D$: $DR=DS$

Now $AB+CD=(AP+PB)+(CR+RD)=(AS+BQ)+(CQ+DS)$.

Rearranging: $=(AS+DS)+(BQ+CQ)=AD+BC$.

Therefore $AB+CD=AD+BC$. Hence proved.

Let parallelogram $ABCD$ circumscribe a circle, touching $AB,BC,CD,DA$ at $P,Q,R,S$ respectively.

Since tangents from an external point are equal: $AP=AS$, $BP=BQ$, $CR=CQ$, $DR=DS$.

Adding all four: $AP+BP+CR+DR=AS+BQ+CQ+DS$, i.e. $(AP+PB)+(CR+RD)=(AS+SD)+(BQ+QC)$, giving

$AB+CD=AD+BC. \quad (1)$

But $ABCD$ is a parallelogram, so opposite sides are equal: $AB=CD$ and $AD=BC. \quad (2)$

Substituting (2) in (1): $2AB=2BC\Rightarrow AB=BC$.

Thus two adjacent sides are equal, and since opposite sides are already equal, all four sides are equal: $AB=BC=CD=DA$.

A parallelogram with all sides equal is a rhombus. Hence proved.

Let the incircle touch $BC$ at $D$, $CA$ at $E$ and $AB$ at $F$. Using equal tangents from each vertex:

• $BD=BF=8$ cm
• $CD=CE=6$ cm
• $AF=AE=x$ (say)

Then $AB=AF+FB=x+8$, $AC=AE+EC=x+6$, and $BC=BD+DC=14$ cm.

Semiperimeter $s=\dfrac{(x+8)+(x+6)+14}{2}=x+14$.

By Heron's formula, with $s-BC=x$, $s-AC=8$, $s-AB=6$:

$\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{(x+14)\cdot x\cdot 6\cdot 8}=\sqrt{48x(x+14)}$.

Also, using the incircle radius $r=4$: $\text{Area}=r\cdot s=4(x+14)$.

Equating and squaring: $16(x+14)^2=48x(x+14)\Rightarrow 16(x+14)=48x\Rightarrow x+14=3x\Rightarrow x=7$.

Therefore $AB=x+8=15$ cm and $AC=x+6=13$ cm.

Given: A circle with centre $O$; $PA$ and $PB$ are tangents from an external point $P$ touching the circle at $A$ and $B$.

To prove: $\angle APB+\angle AOB=180^\circ$.

Proof: Join $OA$ and $OB$. Since a tangent is perpendicular to the radius at the point of contact,

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

Consider quadrilateral $OAPB$. The sum of its interior angles is $360^\circ$:

$\angle OAP+\angle APB+\angle PBO+\angle BOA=360^\circ$.

Substituting the right angles:

$90^\circ+\angle APB+90^\circ+\angle AOB=360^\circ$.

Therefore $\angle APB+\angle AOB=360^\circ-180^\circ=180^\circ$.

Hence the angle between the two tangents is supplementary to the angle subtended by $AB$ at the centre. Proved.

(i) $OA\perp PA$, so triangle $OAP$ is right-angled at $A$. Hence $PA=\sqrt{OP^2-OA^2}=\sqrt{17^2-8^2}=\sqrt{289-64}=\sqrt{225}=15$ m. Each rope is 15 m long (and $PB=PA=15$ m).

(ii) $\angle OAP=90^\circ$, because the tangent at any point of a circle is perpendicular to the radius through the point of contact.

(iii) The quadrilateral $OAPB$ is made up of two congruent right triangles $OAP$ and $OBP$. Area of each $=\dfrac{1}{2}\times OA\times PA=\dfrac{1}{2}\times8\times15=60$ m$^2$.

Total area of $OAPB=2\times60=120$ m$^2$.