Tangents and Intersecting Chords — ICSE Class 10 Maths Important Questions

Answer: (c) $90^{\circ}$90^.

A tangent is perpendicular to the radius at the point of contact.

Answer: (b) $7\,\text{cm}$7\,cm.

The lengths of the two tangents drawn from an external point are equal, so $PB=PA=7\,\text{cm}$PB=PA=7\,cm.

Answer: (a) $8\,\text{cm}$8\,cm.

By the intersecting chords theorem $PA\cdot PB=PC\cdot PD$PA· PB=PC· PD, so $4\times6=3\times PD$4×6=3× PD, giving $PD=\dfrac{24}{3}=8\,\text{cm}$PD=24/3=8\,cm.

Answer: (a) $6\,\text{cm}$6\,cm.

$PB=PA+AB=4+5=9\,\text{cm}$PB=PA+AB=4+5=9\,cm. By the tangent-secant relation $PT^2=PA\cdot PB=4\times9=36$PT^2=PA· PB=4×9=36, so $PT=6\,\text{cm}$PT=6\,cm.

Answer: (c) A is true but R is false.

The intersecting chords relation $PA\cdot PB=PC\cdot PD$PA· PB=PC· PD holds whether the chords meet inside the circle or, when produced, outside it. Question 8 below uses the external case.

The chord of the larger circle touches the smaller circle, so the radius $5\,\text{cm}$5\,cm of the smaller circle is perpendicular to the chord at the point of contact, and it bisects the chord.

Using the right triangle formed with the radius $13\,\text{cm}$13\,cm:
$\text{half-chord}=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\,\text{cm}.$half-chord=√13^2-5^2=√169-25=√144=12\,cm.

Hence the chord $=2\times12=24\,\text{cm}.$=2×12=24\,cm.

The tangent is perpendicular to the radius at the point of contact, so $\angle OTP=90^{\circ}$OTP=90^.

In right triangle $OTP$OTP, by Pythagoras:
$OP=\sqrt{OT^2+PT^2}=\sqrt{8^2+15^2}=\sqrt{64+225}=\sqrt{289}=17\,\text{cm}.$OP=√OT^2+PT^2=√8^2+15^2=√64+225=√289=17\,cm.

When two secants from an external point $P$P cut the circle, $PA\cdot PB=PC\cdot PD$PA· PB=PC· PD.

Here $PB=PA+AB=5+7=12\,\text{cm}$PB=PA+AB=5+7=12\,cm.

$PA\cdot PB=PC\cdot PD$PA· PB=PC· PD
$5\times12=4\times PD\ \Rightarrow\ PD=\frac{60}{4}=15\,\text{cm}.$5×12=4× PD PD=60/4=15\,cm.

Then $CD=PD-PC=15-4=11\,\text{cm}.$CD=PD-PC=15-4=11\,cm.

Since $PA$PA and $PB$PB are tangents, $OA\perp PA$OA PA and $OB\perp PB$OB PB, so $\angle OAP=\angle OBP=90^{\circ}$OAP= OBP=90^.

In quadrilateral $OAPB$OAPB the angles sum to $360^{\circ}$360^:
$\angle AOB=360^{\circ}-90^{\circ}-90^{\circ}-70^{\circ}=110^{\circ}.$AOB=360^-90^-90^-70^=110^.

For $\angle OAB$OAB: the tangents are equal ($PA=PB$PA=PB), so $\triangle PAB$PAB is isosceles and $\angle PAB=\angle PBA=\tfrac12(180^{\circ}-70^{\circ})=55^{\circ}$PAB= PBA=12(180^-70^)=55^. Since $\angle OAP=90^{\circ}$OAP=90^,
$\angle OAB=90^{\circ}-55^{\circ}=35^{\circ}.$OAB=90^-55^=35^.

By the alternate segment theorem, the angle between the tangent $TA$TA and the chord $AB$AB equals the angle in the alternate segment:
$\angle ACB=\angle BAT=62^{\circ}.$ACB= BAT=62^.

For the point $D$D on the minor arc, $ACBD$ACBD is a cyclic quadrilateral, so $\angle ADB$ADB is supplementary to $\angle ACB$ACB:
$\angle ADB=180^{\circ}-62^{\circ}=118^{\circ}.$ADB=180^-62^=118^.

Since the circles touch at $P$P, the point $P$P lies on the line $AB$AB joining the centres. The line through $P$P perpendicular to $AB$AB is therefore perpendicular to both radii $AP$AP and $BP$BP, so it is a tangent to both circles at $P$P.

(i) $TQ$TQ and $TP$TP are tangents from $T$T to the first circle, so $TQ=TP$TQ=TP. $TR$TR and $TP$TP are tangents from $T$T to the second circle, so $TR=TP$TR=TP. Hence $TQ=TR$TQ=TR, and $T$T is the midpoint of $QR$QR.

(ii) In $\triangle TQP$TQP, $TQ=TP$TQ=TP, so $\angle TQP=\angle TPQ$TQP= TPQ. In $\triangle TRP$TRP, $TR=TP$TR=TP, so $\angle TRP=\angle TPR$TRP= TPR.

Adding: $\angle TQP+\angle TRP=\angle TPQ+\angle TPR=\angle QPR$TQP+ TRP= TPQ+ TPR= QPR.

In $\triangle QPR$QPR, $\angle TQP+\angle TRP+\angle QPR=180^{\circ}$TQP+ TRP+ QPR=180^, so $2\angle QPR=180^{\circ}$2 QPR=180^ and $\angle QPR=90^{\circ}$QPR=90^. Hence proved.

Given: Chords $AB$AB and $CD$CD of a circle intersect at a point $P$P inside the circle.

To prove: $PA\cdot PB=PC\cdot PD$PA· PB=PC· PD.

Construction: Join $AC$AC and $BD$BD.

Proof: In triangles $PAC$PAC and $PDB$PDB:

• $\angle APC=\angle DPB$APC= DPB (vertically opposite angles),
• $\angle CAP=\angle BDP$CAP= BDP (angles in the same segment, standing on arc $BC$BC).

By the AA similarity criterion, $\triangle PAC\sim\triangle PDB$PAC PDB.

Hence corresponding sides are proportional:
$\frac{PA}{PD}=\frac{PC}{PB}.$PA/PD=PC/PB.

Cross-multiplying,
$PA\cdot PB=PC\cdot PD.$PA· PB=PC· PD.

This proves the intersecting chords theorem.

Tangents drawn from an external point are equal, so let
$AF=AE=x,\quad BF=BD=y,\quad CD=CE=z.$AF=AE=x, BF=BD=y, CD=CE=z.

Using the three sides:
$AB=AF+FB=x+y=12,$AB=AF+FB=x+y=12,
$BC=BD+DC=y+z=8,$BC=BD+DC=y+z=8,
$CA=CE+EA=z+x=10.$CA=CE+EA=z+x=10.

Add all three equations:
$2(x+y+z)=12+8+10=30\ \Rightarrow\ x+y+z=15.$2(x+y+z)=12+8+10=30 x+y+z=15.

Now subtract each pair:
$x=(x+y+z)-(y+z)=15-8=7\,\text{cm}=AF,$x=(x+y+z)-(y+z)=15-8=7\,cm=AF,
$y=(x+y+z)-(z+x)=15-10=5\,\text{cm}=BD,$y=(x+y+z)-(z+x)=15-10=5\,cm=BD,
$z=(x+y+z)-(x+y)=15-12=3\,\text{cm}=CE.$z=(x+y+z)-(x+y)=15-12=3\,cm=CE.

Hence $AF=7\,\text{cm}$AF=7\,cm, $BD=5\,\text{cm}$BD=5\,cm, $CE=3\,\text{cm}$CE=3\,cm.

(i) By the tangent-secant relation from the external point $P$P:
$PT^2=PA\cdot PB=5\times20=100\ \Rightarrow\ PT=10\,\text{cm}.$PT^2=PA· PB=5×20=100 PT=10\,cm.

(ii) $AB=PB-PA=20-5=15\,\text{cm}.$AB=PB-PA=20-5=15\,cm.

(iii) $OT$OT is a radius to the point of contact, so $OT\perp PT$OT PT and $\angle OTP=90^{\circ}$OTP=90^. In right triangle $OTP$OTP:
$OP=\sqrt{OT^2+PT^2}=\sqrt{24^2+10^2}=\sqrt{576+100}=\sqrt{676}=26\,\text{cm}.$OP=√OT^2+PT^2=√24^2+10^2=√576+100=√676=26\,cm.

(iv) The result $PT^2=PA\cdot PB$PT^2=PA· PB is the tangent-secant property (power of a point): the square of the tangent from an external point equals the product of the whole secant and its external part.