Circles — CBSE Class 10 Maths Important Questions

Answer: (a) $0$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$12 cm

The radius $OP\perp PQ$OP⊥ PQ, so triangle $OPQ$OPQ is right-angled at $P$P. Hence $PQ=\sqrt{OQ^2-OP^2}=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12$PQ=√(OQ²-OP²)=√(13²-5²)=√(169-25)=√144=12 cm.

Answer: (a) $8$8 cm

The chord touches the inner circle, so the perpendicular from the centre (length $=3$=3 cm, the inner radius) bisects the chord. Half the chord $=\sqrt{5^2-3^2}=\sqrt{16}=4$=√(5²-3²)=√16=4 cm, so the chord $=2\times4=8$=2×4=8 cm.

Answer: (a) Both A and R are true and R is the correct explanation of A.

$PA=PB$PA=PB, so $\angle PAB=\angle PBA=\dfrac{180^\circ-70^\circ}{2}=55^\circ$∠ PAB=∠ PBA=(180°-70°)/2=55°. Since $OA\perp PA$OA⊥ PA, $\angle OAP=90^\circ$∠ OAP=90°, so $\angle OAB=90^\circ-55^\circ=35^\circ=\dfrac{1}{2}\times70^\circ$∠ OAB=90°-55°=35°=1/2×70°.

In general $\angle OAB=90^\circ-\left(90^\circ-\dfrac{1}{2}\angle APB\right)=\dfrac{1}{2}\angle APB$∠ OAB=90°-(90°-1/2∠ APB)=1/2∠ APB, which is exactly R.

Let $AB$AB be a diameter of a circle with centre $O$O, $PQ$PQ the tangent at $A$A and $RS$RS the tangent at $B$B.

A tangent is perpendicular to the radius at the point of contact, so $\angle PAB=90^\circ$∠ PAB=90° and $\angle ABS=90^\circ$∠ ABS=90°.

These are alternate angles made by the transversal $AB$AB with the lines $PQ$PQ and $RS$RS, and they are equal. Therefore $PQ\parallel RS$PQ∥ RS. Hence proved.

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

$\tan(\angle APO)=\dfrac{OA}{PA}\Rightarrow \tan30^\circ=\dfrac{3}{PA}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{3}{PA}$tan (∠ APO)=OA/PA⇒ tan 30°=3/PA⇒ 1/√3=3/PA.

Therefore $PA=3\sqrt3$PA=3√3 cm. Each tangent is $3\sqrt3\approx5.20$3√3≈5.20 cm long.

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

To prove: $PA=PB$PA=PB.

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

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

In right triangles $OAP$OAP and $OBP$OBP:

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

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

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

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

Tangents drawn from an external point are equal, so:

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

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

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

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

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

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

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

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

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

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

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

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

Given: A circle with centre $O$O and a tangent $XY$XY to the circle at a point $P$P.

To prove: $OP\perp XY$OP⊥ XY.

Construction: Take any point $Q$Q on $XY$XY other than $P$P and join $OQ$OQ.

Proof: $Q$Q must lie outside the circle. (If $Q$Q lay on or inside the circle, $XY$XY would meet the circle at a second point and would not be a tangent.) So $OQ$OQ meets the circle at a point $R$R between $O$O and $Q$Q.

Hence $OQ>OR$OQ>OR. But $OR=OP$OR=OP (radii of the same circle), so $OQ>OP$OQ>OP.

This is true for every point $Q$Q on $XY$XY except $P$P, so $OP$OP is the shortest of all the distances from $O$O to the points of $XY$XY.

The shortest segment from a point to a line is perpendicular to that line. Therefore $OP\perp XY$OP⊥ XY. Hence proved.

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

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

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

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

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

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

Also, using the incircle radius $r=4$r=4: $\text{Area}=r\cdot s=4(x+14)$Area=r· 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$16(x+14)²=48x(x+14)⇒ 16(x+14)=48x⇒ x+14=3x⇒ x=7.

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

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

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

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

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

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

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

Substituting the right angles:

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

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

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

(i) $OA\perp PA$OA⊥ PA, so triangle $OAP$OAP is right-angled at $A$A. Hence $PA=\sqrt{OP^2-OA^2}=\sqrt{17^2-8^2}=\sqrt{289-64}=\sqrt{225}=15$PA=√(OP²-OA²)=√(17²-8²)=√(289-64)=√225=15 m. Each rope is 15 m long (and $PB=PA=15$PB=PA=15 m).

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

(iii) The quadrilateral $OAPB$OAPB is made up of two congruent right triangles $OAP$OAP and $OBP$OBP. Area of each $=\dfrac{1}{2}\times OA\times PA=\dfrac{1}{2}\times8\times15=60$=1/2× OA× PA=1/2×8×15=60 m$^2$².

Total area of $OAPB=2\times60=120$OAPB=2×60=120 m$^2$².