Tangents and Intersecting Chords — Important Questions

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

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

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

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

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

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

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

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

Answer: (b) Both statements are true. Equality of tangents follows from the RHS congruence of the two right triangles formed, in which the perpendicularity (R) is used; however R by itself is a separate property and is not a full explanation of A, so R is not the correct explanation of A.

The chord of the larger circle touches the smaller circle, so the radius $5\,\text{cm}$ 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 $13\,\text{cm}$:
$\text{half-chord}=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\,\text{cm}.$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Construction: Join $AC$ and $BD$.

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

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

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

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

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

This proves the intersecting chords theorem.

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

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

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

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

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

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

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

(iii) $OT$ is a radius to the point of contact, so $OT\perp PT$ and $\angle OTP=90^{\circ}$. In right triangle $OTP$:
$OP=\sqrt{OT^2+PT^2}=\sqrt{6^2+10^2}=\sqrt{36+100}=\sqrt{136}=2\sqrt{34}\approx11.66\,\text{cm}.$

(iv) The result $PT^2=PA\cdot PB$ 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.