Exercise 10.2 Practice
Tangent Properties: Proofs and Applications
Q1: Tangent Length
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From a point Q, the length of the tangent to a circle is 12 cm and the distance of Q from the centre is 13 cm. The radius of the circle is:
(A) 7 cm (B) 12 cm (C) 5 cm (D) 25 cm
(A) 7 cm (B) 12 cm (C) 5 cm (D) 25 cm
Concept: Tangent is perpendicular to radius. $\angle OPQ = 90^\circ$.
In right $\Delta OPQ$: $OQ^2 = OP^2 + PQ^2$.
$13^2 = r^2 + 12^2 \Rightarrow 169 = r^2 + 144$.
$r^2 = 25 \Rightarrow r = 5$ cm.
Option (C) 5 cm.
Q2: Angle Calculation
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In the figure, if TP and TQ are the two tangents to a circle with centre O so that $\angle POQ = 120^\circ$, then $\angle PTQ$ is equal to:
(A) $60^\circ$ (B) $70^\circ$ (C) $80^\circ$ (D) $90^\circ$
(A) $60^\circ$ (B) $70^\circ$ (C) $80^\circ$ (D) $90^\circ$
In quadrilateral POQT, $\angle OPT = 90^\circ$ and $\angle OQT = 90^\circ$ (Radius $\perp$ Tangent).
Sum of angles = $360^\circ$. So, $\angle PTQ + \angle POQ = 180^\circ$.
$\angle PTQ + 120^\circ = 180^\circ \Rightarrow \angle PTQ = 60^\circ$.
Option (A) $60^\circ$.
Q3: Tangent Inclination
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If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of $60^\circ$, then $\angle POA$ is equal to:
(A) $50^\circ$ (B) $60^\circ$ (C) $70^\circ$ (D) $80^\circ$
(A) $50^\circ$ (B) $60^\circ$ (C) $70^\circ$ (D) $80^\circ$
Triangle OAP $\cong$ Triangle OBP (RHS congruence).
Therefore, $\angle APO = \angle BPO = \frac{1}{2} \angle APB = \frac{1}{2}(60^\circ) = 30^\circ$.
In $\Delta OAP$: $\angle OAP = 90^\circ$.
$\angle POA = 180^\circ - (90^\circ + 30^\circ) = 60^\circ$.
Option (B) $60^\circ$.
Q4: Proof
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Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Let PQ be a diameter. Let tangents AB and CD be drawn at P and Q respectively.
Since radius is perpendicular to tangent, $\angle OP B = 90^\circ$ and $\angle OQC = 90^\circ$.
These two angles are consecutive interior angles (or alternate interior angles depending on view). Since $\angle OPB + \angle OQD = 180^\circ$ (if on same side) or Alternate angles are equal.
Consider transversal PQ intersecting AB and CD. Alternate angles $\angle APQ$ and $\angle PQD$ are both $90^\circ$.
Since alternate interior angles are equal, the lines AB and CD are parallel.
Q5: Proof
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Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Let the tangent be AB at point P on circle with center O.
We know that radius OP $\perp$ AB. So $\angle OPA = 90^\circ$.
If there is another point O' such that O'P is perpendicular to AB, then $\angle O'PA = 90^\circ$.
This implies $\angle OPA = \angle O'PA$, which is only possible if O and O' lie on the same line.
Thus, the perpendicular at the point of contact must pass through the centre.
Q6: Pythagorean Triple
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The length of a tangent from a point A at distance 10 cm from the centre of the circle is 8 cm. Find the radius of the circle.
Let O be center, T be point of contact. OA = 10 cm, AT = 8 cm.
$\Delta OTA$ is a right triangle at T.
$OT^2 + AT^2 = OA^2 \Rightarrow r^2 + 8^2 = 10^2$.
$r^2 = 100 - 64 = 36 \Rightarrow r = 6$ cm.
Radius is 6 cm.
Q7: Concentric Circles
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Two concentric circles are of radii 13 cm and 5 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Let AB be the chord. It touches smaller circle at P. OP $\perp$ AB.
In right $\Delta APO$: $AO^2 = AP^2 + OP^2$.
$13^2 = AP^2 + 5^2 \Rightarrow 169 = AP^2 + 25$.
$AP^2 = 144 \Rightarrow AP = 12$ cm.
Total length AB = $2 \times AP = 2 \times 12 = 24$ cm.
Length of chord is 24 cm.
Q8: Quadrilateral Property
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A quadrilateral ABCD is drawn to circumscribe a circle. Prove that $AB + CD = AD + BC$.
Tangents from an external point are equal.
$AP = AS$, $BP = BQ$, $CR = CQ$, $DR = DS$.
Add LHS and RHS: $(AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ)$.
$AB + CD = AD + BC$.
Hence Proved.
Q9: Parallel Tangents
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In a circle with centre O, XY and X'Y' are two parallel tangents, and another tangent AB with point of contact C intersects XY at A and X'Y' at B. Prove that $\angle AOB = 90^\circ$.
Join OC.
$\Delta OPA \cong \Delta OCA$ (RHS). So $\angle POA = \angle COA$.
Similarly $\angle QOB = \angle COB$.
POQ is a diameter (straight line), so $2\angle COA + 2\angle COB = 180^\circ$.
$2(\angle COA + \angle COB) = 180^\circ \Rightarrow \angle AOB = 90^\circ$.
Hence Proved.
Q10: Supplementary Angles
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Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Let tangents be PA and PB. O is center.
In quadrilateral OAPB, $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$.
Sum of angles = 360°. So $\angle APB + \angle AOB + 90 + 90 = 360$.
$\angle APB + \angle AOB = 180^\circ$.
Hence angles are supplementary.
Q11: Rhombus Proof
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Prove that the parallelogram circumscribing a circle is a rhombus.
Let ABCD be the parallelogram. From Q8, we know $AB + CD = AD + BC$.
Since it's a parallelogram, $AB = CD$ and $AD = BC$.
So, $2AB = 2AD \Rightarrow AB = AD$.
Since adjacent sides are equal, ABCD is a rhombus.
Hence Proved.
Q12: Triangle Area
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A triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 4 cm and 3 cm respectively. Find the sides AB and AC.
$BD=4, DC=3$. Tangents equal: $BE=4, CF=3$. Let $AE=AF=x$.
Sides: $a=7, b=3+x, c=4+x$. $s = \frac{7+3+x+4+x}{2} = 7+x$.
Area $\Delta ABC = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(7+x)(x)(4)(3)} = \sqrt{12x(7+x)}$.
Area also = $\frac{1}{2}r(P) = \frac{1}{2}(2)(14+2x) = 14+2x$.
Square both sides: $12x(7+x) = (2(7+x))^2 = 4(7+x)^2$.
$3x = 7+x \Rightarrow 2x = 7 \Rightarrow x = 3.5$.
$3x = 7+x \Rightarrow 2x = 7 \Rightarrow x = 3.5$.
$AB = 4 + 3.5 = 7.5$ cm, $AC = 3 + 3.5 = 6.5$ cm.
Sides are 7.5 cm and 6.5 cm.
Q13: Supplementary Angles
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Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Join center O to all points of contact and vertices.
We get pairs of congruent triangles (e.g., $\Delta OAP \cong \Delta OAS$). So adjacent angles at center are equal (say a, a, b, b, c, c, d, d).
$2(a+b+c+d) = 360^\circ \Rightarrow a+b+c+d = 180^\circ$.
Angle by AB = $a+b$, Angle by CD = $c+d$. Sum is $180^\circ$.
Hence Proved.