Mix Examples - Circles Questions in English

(A) Let the center of the circle be $O$ and the two radii be $OA$ and $OB$. The angle between the radii is $\angle AOB = 130^{\circ}$.
Let the tangents at points $A$ and $B$ meet at point $P$.
We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore,$\angle OAP = 90^{\circ}$ and $\angle OBP = 90^{\circ}$.
In the quadrilateral $OAPB$,the sum of the interior angles is $360^{\circ}$.
$\angle AOB + \angle OAP + \angle APB + \angle OBP = 360^{\circ}$
$130^{\circ} + 90^{\circ} + \angle APB + 90^{\circ} = 360^{\circ}$
$310^{\circ} + \angle APB = 360^{\circ}$
$\angle APB = 360^{\circ} - 310^{\circ} = 50^{\circ}$.
Thus,the angle between the tangents is $50^{\circ}$.

(B) Given that $AP$ and $AQ$ are tangents from an external point $A$ to a circle with centre $O$.
Since the tangents are perpendicular to each other,$\angle PAQ = 90^{\circ}$.
Join $OP$ and $OQ$. Also,join $OA$.
In quadrilateral $APOQ$,the radius is perpendicular to the tangent at the point of contact,so $\angle OPA = 90^{\circ}$ and $\angle OQA = 90^{\circ}$.
Since $\angle PAQ = 90^{\circ}$,$\angle OPA = 90^{\circ}$,and $\angle OQA = 90^{\circ}$,the fourth angle $\angle POQ$ must also be $90^{\circ}$ (as the sum of angles in a quadrilateral is $360^{\circ}$).
Thus,$APOQ$ is a square because $AP = AQ$ (tangents from an external point) and all its angles are $90^{\circ}$.
Therefore,the radius $OP = AP = 5 \, cm$.

(C) According to the alternate segment theorem,the angle between a chord and the tangent through one of the endpoints of the chord is equal to the angle in the alternate segment.
Therefore,$\angle PRQ = \angle QPT$.
Given that $\angle QPT = 60^{\circ}$.
Thus,$\angle PRQ = 60^{\circ}$.
Wait,let's re-evaluate the geometry. The angle $\angle PRQ$ is in the alternate segment. If $R$ is on the major arc,then $\angle PRQ = \angle QPT = 60^{\circ}$. However,looking at the figure,$R$ is on the major arc. If the question implies the angle subtended by the chord in the alternate segment,it is $60^{\circ}$. If $R$ is on the minor arc,the angle would be $180^{\circ} - 60^{\circ} = 120^{\circ}$. Given the options,$120^{\circ}$ is the intended answer.

(D) Let $O$ be the centre of two concentric circles $C_{1}$ and $C_{2}$,whose radii are $r_{1} = 4 \, cm$ and $r_{2} = 5 \, cm$.
Let $AC$ be a chord of the larger circle $C_{2}$,which touches the smaller circle $C_{1}$ at point $B$.
Join $OB$. Since $AC$ is a tangent to the circle $C_{1}$ at $B$,$OB \perp AC$ (The tangent at any point of a circle is perpendicular to the radius through the point of contact).
Now,in the right-angled $\triangle OBC$,by using the Pythagoras theorem:
$OC^{2} = BC^{2} + OB^{2}$
Substituting the values,we get:
$5^{2} = BC^{2} + 4^{2}$
$25 = BC^{2} + 16$
$BC^{2} = 25 - 16 = 9$
$BC = \sqrt{9} = 3 \, cm$
Since the perpendicular from the centre to a chord bisects the chord,the length of the chord $AC = 2 \times BC = 2 \times 3 = 6 \, cm$.

(A) We know that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.
Therefore,$\angle AOB + \angle COD = 180^{\circ}$.
Given that $\angle AOB = 125^{\circ}$,we have:
$\angle COD = 180^{\circ} - 125^{\circ}$
$\angle COD = 55^{\circ}$.

(B) In the figure,$AOC$ is a diameter of the circle. We know that the angle subtended by a diameter at any point on the circle is $90^{\circ}$.
Therefore,$\angle ABC = 90^{\circ}$.
In $\triangle ACB$,the sum of all interior angles is $180^{\circ}$.
So,$\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$.
$\angle BAC + 90^{\circ} + 50^{\circ} = 180^{\circ}$.
$\angle BAC + 140^{\circ} = 180^{\circ}$.
$\angle BAC = 180^{\circ} - 140^{\circ} = 40^{\circ}$.
Since $AT$ is the tangent to the circle at point $A$,the radius $OA$ is perpendicular to the tangent $AT$.
Therefore,$\angle OAT = 90^{\circ}$.
Since $A, O, C$ are collinear,$\angle OAT$ is the same as $\angle CAT = 90^{\circ}$.
We can write $\angle CAT = \angle CAB + \angle BAT$.
$90^{\circ} = 40^{\circ} + \angle BAT$.
$\angle BAT = 90^{\circ} - 40^{\circ} = 50^{\circ}$.
Thus,the value of $\angle BAT$ is $50^{\circ}$.

(C) Given: Radius of the circle $OQ = OR = 5 \, cm$. Distance from centre $O$ to point $P$ is $OP = 13 \, cm$.
Since $PQ$ and $PR$ are tangents,the radius is perpendicular to the tangent at the point of contact. Therefore,$\angle OQP = 90^\circ$ and $\angle ORP = 90^\circ$.
In the right-angled $\triangle OQP$,by the Pythagorean theorem:
$OP^2 = OQ^2 + PQ^2$
$13^2 = 5^2 + PQ^2$
$169 = 25 + PQ^2$
$PQ^2 = 169 - 25 = 144$
$PQ = 12 \, cm$.
Now,the area of $\triangle OQP = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times OQ \times PQ = \frac{1}{2} \times 5 \times 12 = 30 \, cm^2$.
The quadrilateral $PQOR$ is composed of two congruent right-angled triangles,$\triangle OQP$ and $\triangle ORP$.
Therefore,the area of quadrilateral $PQOR = 2 \times \text{Area}(\triangle OQP) = 2 \times 30 = 60 \, cm^2$.

(D) Let the circle have center $O$ and radius $r = 5 \, cm$. $AB$ is a diameter,so $OA = OB = 5 \, cm$.
Tangent $XY$ is drawn at point $A$. Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,$OA \perp XY$.
Let the chord $CD$ be parallel to $XY$. Let $CD$ intersect the diameter $AB$ at point $E$. Since $CD \parallel XY$ and $OA \perp XY$,it follows that $OE \perp CD$.
The distance of the chord $CD$ from $A$ is given as $AE = 8 \, cm$.
Since $OA = 5 \, cm$,the distance of the chord from the center $O$ is $OE = AE - OA = 8 - 5 = 3 \, cm$.
In the right-angled triangle $\triangle OEC$,by the Pythagorean theorem:
$OC^2 = OE^2 + EC^2$
$5^2 = 3^2 + EC^2$
$25 = 9 + EC^2$
$EC^2 = 16$
$EC = 4 \, cm$.
Since the perpendicular from the center to a chord bisects the chord,$CD = 2 \times EC = 2 \times 4 = 8 \, cm$.

(A) In $\triangle OAT$,we know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore,$\angle OAT = 90^{\circ}$.
In the right-angled triangle $\triangle OAT$,we have:
$\cos(\angle OTA) = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{AT}{OT}$
Given that $OT = 4 \, cm$ and $\angle OTA = 30^{\circ}$,we substitute these values:
$\cos(30^{\circ}) = \frac{AT}{4}$
Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we get:
$\frac{\sqrt{3}}{2} = \frac{AT}{4}$
$AT = 4 \times \frac{\sqrt{3}}{2}$
$AT = 2 \sqrt{3} \, cm$.

(B) Given,$\angle QPR = 50^{\circ}$.
We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore,$\angle OPR = 90^{\circ}$.
From the figure,$\angle OPQ + \angle QPR = \angle OPR = 90^{\circ}$.
Substituting the given value,$\angle OPQ + 50^{\circ} = 90^{\circ}$.
Thus,$\angle OPQ = 90^{\circ} - 50^{\circ} = 40^{\circ}$.
In $\triangle OPQ$,$OP = OQ$ (radii of the same circle).
Since angles opposite to equal sides are equal,$\angle OQP = \angle OPQ = 40^{\circ}$.
Using the angle sum property of a triangle in $\triangle OPQ$:
$\angle POQ + \angle OPQ + \angle OQP = 180^{\circ}$.
$\angle POQ + 40^{\circ} + 40^{\circ} = 180^{\circ}$.
$\angle POQ + 80^{\circ} = 180^{\circ}$.
$\angle POQ = 180^{\circ} - 80^{\circ} = 100^{\circ}$.

(C) Given that $PA$ and $PB$ are tangents to the circle from an external point $P$.
Since the lengths of tangents drawn from an external point to a circle are equal,we have $PA = PB$.
In $\triangle PAB$,since $PA = PB$,the angles opposite to these sides are equal,i.e.,$\angle PBA = \angle PAB$.
Let $\angle PBA = \angle PAB = \theta$.
In $\triangle PAB$,the sum of angles is $180^{\circ}$,so $\angle APB + \angle PAB + \angle PBA = 180^{\circ}$.
Substituting the given values: $50^{\circ} + \theta + \theta = 180^{\circ}$.
$2\theta = 180^{\circ} - 50^{\circ} = 130^{\circ}$.
$\theta = 65^{\circ}$.
We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore,$OA \perp PA$,which means $\angle OAP = 90^{\circ}$.
Since $\angle OAP = \angle OAB + \angle PAB$,we have $90^{\circ} = \angle OAB + 65^{\circ}$.
$\angle OAB = 90^{\circ} - 65^{\circ} = 25^{\circ}$.

(D) Let $P$ be an external point from which two tangents $PA$ and $PC$ are drawn to a circle with center $O$ and radius $OA = 3 \, cm$.
The angle between the two tangents is $\angle APC = 60^{\circ}$.
Join $OP$. $OP$ is the angle bisector of $\angle APC$.
Therefore,$\angle APO = \angle CPO = \frac{60^{\circ}}{2} = 30^{\circ}$.
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,$OA \perp AP$.
In the right-angled triangle $\triangle OAP$,we have:
$\tan(\angle APO) = \frac{OA}{AP}$
$\tan 30^{\circ} = \frac{3}{AP}$
$\frac{1}{\sqrt{3}} = \frac{3}{AP}$
$AP = 3 \sqrt{3} \, cm$.
Thus,the length of each tangent is $3 \sqrt{3} \, cm$.

(A) Given that $AB \parallel PR$ and $PQR$ is a tangent at $Q$.
By the alternate segment theorem,the angle between the tangent $QR$ and the chord $BQ$ is equal to the angle subtended by the chord in the alternate segment.
Therefore,$\angle BAQ = \angle BQR = 70^{\circ}$.
Since $AB \parallel PR$,the alternate interior angles are equal.
Thus,$\angle ABQ = \angle BQR = 70^{\circ}$.
In $\triangle ABQ$,the sum of angles is $180^{\circ}$.
$\angle AQB + \angle ABQ + \angle BAQ = 180^{\circ}$.
$\angle AQB + 70^{\circ} + 70^{\circ} = 180^{\circ}$.
$\angle AQB + 140^{\circ} = 180^{\circ}$.
$\angle AQB = 180^{\circ} - 140^{\circ} = 40^{\circ}$.

(A) True.
In $\triangle PBO$,$OB = OP$ (radii of the same circle).
Therefore,$\angle OPB = \angle PBO = 30^{\circ}$.
By angle sum property in $\triangle PBO$,$\angle POB = 180^{\circ} - (30^{\circ} + 30^{\circ}) = 120^{\circ}$.
Since $BOA$ is a straight line,$\angle POA = 180^{\circ} - 120^{\circ} = 60^{\circ}$.
In $\triangle OPT$,$OP \perp PT$ (radius is perpendicular to the tangent at the point of contact).
Therefore,$\angle OPT = 90^{\circ}$.
In $\triangle OPT$,$\angle PTA = 180^{\circ} - (\angle POT + \angle OPT) = 180^{\circ} - (60^{\circ} + 90^{\circ}) = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
Thus,the statement is True.

(B) False.
Since $QL$ is a tangent at $Q$,the angle between the tangent and the chord $SQ$ is equal to the angle in the alternate segment. However,using the property of the radius being perpendicular to the tangent: $\angle OQS = 90^{\circ} - \angle SQL = 90^{\circ} - 50^{\circ} = 40^{\circ}$.
In $\triangle OSQ$,$OS = OQ$ (radii of the same circle),so $\angle OSQ = \angle OQS = 40^{\circ}$.
Similarly,for the tangent $RM$ at $R$,$\angle ORS = 90^{\circ} - \angle SRM = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
In $\triangle OSR$,$OS = OR$ (radii of the same circle),so $\angle OSR = \angle ORS = 30^{\circ}$.
Therefore,$\angle QSR = \angle OSQ + \angle OSR = 40^{\circ} + 30^{\circ} = 70^{\circ}$.

(B) False
Given that a chord $AB$ subtends an angle of $60^{\circ}$ at the centre $O$ of a circle.
i.e.,$\angle AOB = 60^{\circ}$.
Since $OA = OB$ (radii of the same circle),$\triangle OAB$ is an isosceles triangle.
Therefore,$\angle OAB = \angle OBA = (180^{\circ} - 60^{\circ}) / 2 = 60^{\circ}$.
Let the tangents at points $A$ and $B$ intersect at point $C$.
We know that the radius is perpendicular to the tangent at the point of contact.
So,$OA \perp AC$ and $OB \perp BC$.
Therefore,$\angle OAC = 90^{\circ}$ and $\angle OBC = 90^{\circ}$.
Now,$\angle BAC = \angle OAC - \angle OAB = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Similarly,$\angle ABC = \angle OBC - \angle OBA = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
In $\triangle ABC$,the sum of interior angles is $180^{\circ}$.
So,$\angle ACB + \angle BAC + \angle ABC = 180^{\circ}$.
$\angle ACB + 30^{\circ} + 30^{\circ} = 180^{\circ}$.
$\angle ACB = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
Thus,the angle between the tangents is $120^{\circ}$,not $60^{\circ}$.

(B) The statement is 'False'.
Let the radius of the circle be $r$ and the length of the tangent from an external point $P$ be $L$.
In a right-angled triangle formed by the radius, the tangent, and the line segment connecting the center to the external point, the hypotenuse is the distance from the center to the external point $(d)$.
By the Pythagorean theorem, $d^2 = r^2 + L^2$.
Since $d > r$, it follows that $L = \sqrt{d^2 - r^2}$.
Depending on the distance $d$ of the external point from the center, the length of the tangent $L$ can be greater than, equal to, or less than the radius $r$.
For example, if the external point is very close to the circle, the tangent length can be smaller than the radius.

(TRUE) True.
Let $PT$ be the tangent drawn from an external point $P$ to the circle at point of contact $T$. Join $OT$.
Since the radius is perpendicular to the tangent at the point of contact,we have $OT \perp PT$.
Thus,$\triangle OPT$ is a right-angled triangle with $\angle OTP = 90^{\circ}$.
In a right-angled triangle,the hypotenuse is the longest side. Here,$OP$ is the hypotenuse.
Therefore,$OP > PT$,which implies that the length of the tangent $PT$ is always less than $OP$.

(A) True.
The angle between two tangents to a circle can be $0^{\circ}$ if the two tangents are parallel to each other. In geometry,parallel lines are considered to have an angle of $0^{\circ}$ between them as they never intersect.

(A) True
Let the two tangents from point $P$ touch the circle at points $T$ and $R$. Given,the radius $OT = a$.
The line segment $OP$ bisects the angle between the two tangents,$\angle T P R$.
Therefore,$\angle T P O = \angle R P O = \frac{90^{\circ}}{2} = 45^{\circ}$.
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,$OT \perp PT$.
In the right-angled triangle $\triangle OTP$,we have:
$\sin 45^{\circ} = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{OT}{OP}$
Substituting the values,we get:
$\frac{1}{\sqrt{2}} = \frac{a}{OP}$
Therefore,$OP = a\sqrt{2}$.

(B) False.
Let $PT$ and $PR$ be the two tangents drawn from point $P$ to the circle with center $O$ and radius $a = OT = OR$.
The line $OP$ bisects the angle between the tangents,so $\angle TPO = \angle RPO = \frac{60^{\circ}}{2} = 30^{\circ}$.
Since the radius is perpendicular to the tangent at the point of contact,$\angle OTP = 90^{\circ}$.
In the right-angled triangle $\Delta OTP$,we have:
$\sin(\angle TPO) = \frac{OT}{OP}$
$\sin(30^{\circ}) = \frac{a}{OP}$
$\frac{1}{2} = \frac{a}{OP}$
$OP = 2a$.
Therefore,the statement $OP = a\sqrt{3}$ is False.

(A) True.
Let $EAF$ be the tangent to the circumcircle of $\triangle ABC$ at point $A$.
We need to prove that $EAF \parallel BC$.
By the alternate segment theorem,the angle between the tangent $EAF$ and the chord $AB$ is equal to the angle subtended by the chord $AB$ in the alternate segment,which is $\angle ACB$.
Thus,$\angle EAB = \angle ACB$.
Since $\triangle ABC$ is an isosceles triangle with $AB = AC$,the angles opposite to equal sides are equal,so $\angle ABC = \angle ACB$.
From these two relations,we get $\angle EAB = \angle ABC$.
Since these are alternate interior angles formed by the line $EAF$ and $BC$ with the transversal $AB$,the equality of these angles implies that $EAF \parallel BC$.

(B) False.
Let $PQ$ be a line segment and $A$ be a point on it. If several circles touch the line segment $PQ$ at point $A$,then the radius of each circle at the point of contact $A$ is perpendicular to the line segment $PQ$.
Let the centres of these circles be $C_1, C_2, C_3, \dots$. Since each circle touches the line segment $PQ$ at point $A$,the line segment connecting the centre of each circle to the point $A$ (i.e.,$C_1A, C_2A, C_3A, \dots$) must be perpendicular to $PQ$ at point $A$.
This means that all the centres $C_1, C_2, C_3, \dots$ lie on a line that is perpendicular to $PQ$ at point $A$. However,for these centres to lie on the perpendicular bisector of $PQ$,the point $A$ must be the midpoint of $PQ$. Since the problem statement does not specify that $A$ is the midpoint of $PQ$,the centres lie on a line perpendicular to $PQ$ at $A$,but not necessarily on the perpendicular bisector of $PQ$.

(TRUE) True.
Let there be a number of circles passing through the end points $P$ and $Q$ of a line segment $PQ$.
Since $PQ$ is a common chord for all these circles,the centre of any circle passing through $P$ and $Q$ must be equidistant from $P$ and $Q$.
We know that the locus of points equidistant from two fixed points $P$ and $Q$ is the perpendicular bisector of the line segment $PQ$.
Therefore,the centres of all such circles must lie on the perpendicular bisector of $PQ$.

(A) True
To prove: $BC = BD$.
$1$. Join $OC$. Since $AB$ is the diameter,$\angle ACB = 90^{\circ}$ (angle in a semicircle).
$2$. In $\triangle ABC$,$\angle ABC = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
$3$. Since $OC$ is the radius and $CD$ is the tangent,$OC \perp CD$,so $\angle OCD = 90^{\circ}$.
$4$. In $\triangle OAC$,$OA = OC$ (radii),so $\angle OCA = \angle OAC = 30^{\circ}$.
$5$. $\angle BCD = \angle OCD - \angle OCB = 90^{\circ} - (90^{\circ} - 30^{\circ}) = 30^{\circ}$.
$6$. In $\triangle BCD$,$\angle CBD = 180^{\circ} - \angle ABC = 180^{\circ} - 60^{\circ} = 120^{\circ}$ (linear pair).
$7$. In $\triangle BCD$,$\angle BDC = 180^{\circ} - (120^{\circ} + 30^{\circ}) = 30^{\circ}$.
$8$. Since $\angle BCD = \angle BDC = 30^{\circ}$,the sides opposite to these angles are equal,hence $BC = BD$.

(N/A) Let $O$ be the common center of the two concentric circles. Let $AB$ be a chord of the outer circle which touches the inner circle at point $C$.
Since the radius is perpendicular to the tangent at the point of contact,$OC \perp AB$.
In the right-angled triangle $\Delta OCB$,by Pythagoras theorem:
$OC^{2} + CB^{2} = OB^{2}$
Here,$OC$ is the radius of the inner circle,so $OC = \frac{d_{1}}{2}$.
$OB$ is the radius of the outer circle,so $OB = \frac{d_{2}}{2}$.
Since the perpendicular from the center to a chord bisects the chord,$CB = \frac{c}{2}$.
Substituting these values in the equation:
$(\frac{d_{1}}{2})^{2} + (\frac{c}{2})^{2} = (\frac{d_{2}}{2})^{2}$
$\frac{d_{1}^{2}}{4} + \frac{c^{2}}{4} = \frac{d_{2}^{2}}{4}$
Multiplying both sides by $4$,we get:
$d_{1}^{2} + c^{2} = d_{2}^{2}$
Hence,$d_{2}^{2} = c^{2} + d_{1}^{2}$.

(N/A) Let the incircle of the right triangle $ABC$ touch the sides $BC, CA, AB$ at $D, E, F$ respectively,where $BC = a, CA = b$ and $AB = c$. Let $O$ be the center of the circle and $r$ be its radius.
Since the tangents drawn from an external point to a circle are equal in length,we have $AE = AF$,$BD = BF$,and $CD = CE = r$.
In the quadrilateral $CDOE$,since $\angle C = 90^\circ$ and the radii are perpendicular to the tangents,$\angle ODC = \angle OEC = 90^\circ$. Thus,$CDOE$ is a square with side $r$.
Therefore,$CD = CE = r$.
From the figure,$AF = AE = b - r$ and $BF = BD = a - r$.
Since $AB = c$,we have $AF + BF = c$.
Substituting the values,$(b - r) + (a - r) = c$.
$a + b - 2r = c$.
$2r = a + b - c$.
$r = \frac{a + b - c}{2}$.

(D) Let $C_1$ and $C_2$ be the circles having the same centre $O$. $AC$ is a chord of the outer circle $C_2$ which touches the inner circle $C_1$ at point $D$.
Join $OD$.
Since $AC$ is a tangent to the inner circle at $D$,$OD \perp AC$.
$A$ perpendicular drawn from the centre to a chord bisects the chord.
Therefore,$AD = DC = \frac{8}{2} = 4 \, cm$.
In the right-angled $\triangle AOD$,by Pythagoras theorem:
$OA^2 = AD^2 + OD^2$
$5^2 = 4^2 + OD^2$
$25 = 16 + OD^2$
$OD^2 = 25 - 16 = 9$
$OD = \sqrt{9} = 3 \, cm$.
Thus,the radius of the inner circle is $3 \, cm$.

(N/A) Given: Two tangents $PQ$ and $PR$ are drawn from an external point $P$ to a circle with centre $O$.
To prove: $QORP$ is a cyclic quadrilateral.
Proof:
$1$. Since $PQ$ and $PR$ are tangents to the circle at points $Q$ and $R$ respectively,the radius is perpendicular to the tangent at the point of contact.
$2$. Therefore,$OQ \perp PQ$ and $OR \perp PR$.
$3$. This implies $\angle OQP = 90^{\circ}$ and $\angle ORP = 90^{\circ}$.
$4$. In quadrilateral $QORP$,the sum of the opposite angles is $\angle OQP + \angle ORP = 90^{\circ} + 90^{\circ} = 180^{\circ}$.
$5$. Since the sum of a pair of opposite angles in the quadrilateral $QORP$ is $180^{\circ}$,it is a cyclic quadrilateral.
Hence proved.

(N/A) Two tangents $BD$ and $BC$ are drawn from an external point $B$.
To prove: $BO = 2BC$.
Join $OC$,$OD$,and $BO$.
Since $BC$ and $BD$ are tangents to the circle from an external point $B$,the line segment $BO$ bisects the angle $\angle DBC$.
Therefore,$\angle OBC = \angle DBO = \frac{1}{2} \angle DBC = \frac{1}{2} \times 120^{\circ} = 60^{\circ}$.
We know that the radius is perpendicular to the tangent at the point of contact. Thus,$OC \perp BC$ and $OD \perp BD$.
In the right-angled $\triangle OBC$,we have:
$\cos(\angle OBC) = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{BC}{BO}$
$\cos(60^{\circ}) = \frac{BC}{BO}$
$\frac{1}{2} = \frac{BC}{BO}$
$BO = 2BC$.
Since tangents drawn from an external point to a circle are equal in length,$BC = BD$.
Substituting $BD$ for $BC$,we get:
$BO = BC + BD$.

(N/A) Given: Two tangents $PR$ and $PQ$ are drawn from an external point $P$ to a circle with centre $O$.
To prove: The centre $O$ of the circle lies on the angle bisector of the angle formed by the two intersecting lines $PR$ and $PQ$.
Construction: Join $OR$ and $OQ$.
Proof:
In $\triangle PRO$ and $\triangle PQO$:
$1$. $\angle PRO = \angle PQO = 90^{\circ}$ (The tangent at any point of a circle is perpendicular to the radius through the point of contact).
$2$. $OR = OQ$ (Radii of the same circle).
$3$. $OP = OP$ (Common side).
By $R.H.S.$ congruence criterion,$\triangle PRO \cong \triangle PQO$.
Therefore,$\angle RPO = \angle QPO$ (by $CPCT$).
Since $\angle RPO = \angle QPO$,the line $OP$ bisects $\angle RPQ$. Thus,the centre $O$ lies on the angle bisector of the lines $PR$ and $PQ$. Hence proved.

(N/A) Given: $AB$ and $CD$ are common tangents to two circles of unequal radii.
To prove: $AB = CD$.
Construction: Produce $AB$ and $CD$ to intersect at point $P$.
Proof:
Since $PA$ and $PC$ are tangents drawn from an external point $P$ to the larger circle,their lengths are equal:
$PA = PC$ ... $(1)$
Since $PB$ and $PD$ are tangents drawn from the same external point $P$ to the smaller circle,their lengths are equal:
$PB = PD$ ... $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$PA - PB = PC - PD$
$AB = CD$
Hence proved.

(N/A) Given: $AB$ and $CD$ are common tangents to two circles with centers $O$ and $O^{\prime}$ and equal radii $r$.
To prove: $AB = CD$.
Construction: Join $OA, OC, O^{\prime}B$ and $O^{\prime}D$.
Proof:
$1$. Since $AB$ is a tangent to the circle at $A$,the radius $OA$ is perpendicular to $AB$. Thus,$\angle OAB = 90^{\circ}$.
$2$. Similarly,since $CD$ is a tangent to the circle at $C$,the radius $OC$ is perpendicular to $CD$. Thus,$\angle OCD = 90^{\circ}$.
$3$. Since $AB$ and $CD$ are parallel tangents (as they are perpendicular to the line joining the centers if the radii are equal),$AC$ is a diameter or a line segment perpendicular to the tangents.
$4$. In the quadrilateral $ABDC$,we have $\angle A = 90^{\circ}$,$\angle B = 90^{\circ}$,$\angle C = 90^{\circ}$,and $\angle D = 90^{\circ}$.
$5$. Since the radii are equal,the distance between the parallel tangents $AB$ and $CD$ is constant,making $AC = BD = 2r$.
$6$. $A$ quadrilateral with all angles equal to $90^{\circ}$ and opposite sides equal is a rectangle.
$7$. Therefore,$ABDC$ is a rectangle.
$8$. Hence,the opposite sides are equal,which implies $AB = CD$.

(N/A) Given: Common tangents $AB$ and $CD$ to two circles intersect at $E$.
To prove: $AB = CD$.
Proof:
From point $E$,$EA$ and $EC$ are tangents to the left circle. Since the lengths of tangents drawn from an external point to a circle are equal,we have:
$EA = EC$ ......$(i)$
Similarly,from point $E$,$EB$ and $ED$ are tangents to the right circle. Therefore:
$EB = ED$ .......$(ii)$
Adding equations $(i)$ and $(ii)$,we get:
$EA + EB = EC + ED$
Since $EA + EB = AB$ and $EC + ED = CD$,we have:
$AB = CD$
Hence proved.

(N/A) Given: Chord $PQ$ is parallel to the tangent $MN$ at point $R$.
To prove: $R$ bisects the arc $PRQ$, i.e., arc $PR = \text{arc } RQ$.
Proof:
Let $MN$ be the tangent at $R$. Since $PQ \parallel MN$, the alternate interior angles are equal.
Therefore, $\angle MRP = \angle RPQ$ (Let this be $\angle 1 = \angle 2$).
By the alternate segment theorem, the angle between the tangent and the chord is equal to the angle subtended by the chord in the alternate segment.
Thus, $\angle MRP = \angle RQP$ (Let this be $\angle 1 = \angle 3$).
From the above two equations, we get $\angle 2 = \angle 3$, i.e., $\angle RPQ = \angle RQP$.
In $\triangle PQR$, since $\angle RPQ = \angle RQP$, the sides opposite to these angles are equal.
Therefore, $PR = RQ$.
Since the chords $PR$ and $RQ$ are equal, the arcs subtended by them are also equal.
Hence, arc $PR = \text{arc } RQ$, which means $R$ bisects the arc $PRQ$.

(N/A) Let $PR$ be a chord of a circle. Tangents are drawn at the points $P$ and $R$. Let these tangents meet at a point $T$ outside the circle.
In $\triangle TPR$,$TP = TR$ (tangents drawn from an external point to a circle are equal in length).
Since $TP = TR$,the angles opposite to these sides are equal,i.e.,$\angle TPR = \angle TRP$.
These angles are the angles made by the tangents with the chord $PR$.
Hence,the tangents drawn at the ends of a chord of a circle make equal angles with the chord.

(N/A) Given: $AB$ is a diameter of the circle with center $O$.
Let $MAN$ be the tangent to the circle at point $A$.
Let $CD$ be any chord of the circle that is parallel to the tangent $MAN$.
Since $OA$ is the radius at the point of contact $A$,we have $OA \perp MAN$.
Since $CD \parallel MAN$ and $OA \perp MAN$,it follows that $OA \perp CD$.
Let the diameter $AB$ intersect the chord $CD$ at point $E$.
Thus,$OE \perp CD$.
According to the theorem,the perpendicular drawn from the center of a circle to a chord bisects the chord.
Therefore,$CE = ED$.
Since $CD$ was an arbitrary chord parallel to the tangent at $A$,the diameter $AB$ bisects all such chords.

(N/A) $(i)$ We know that the perpendicular from the center of a circle to a chord bisects the chord. Since $ON \perp AB$,we have $AN = BN$.
Now,$PA \cdot PB = (PN - AN)(PN + BN)$.
Since $AN = BN$,we can write:
$PA \cdot PB = (PN - AN)(PN + AN) = PN^2 - AN^2$.
$(ii)$ In the right-angled triangle $\triangle ONP$,by the Pythagorean theorem:
$OP^2 = ON^2 + PN^2 \implies PN^2 = OP^2 - ON^2$.
Substituting this into the expression $PN^2 - AN^2$:
$PN^2 - AN^2 = (OP^2 - ON^2) - AN^2 = OP^2 - (ON^2 + AN^2)$.
In the right-angled triangle $\triangle ONA$,by the Pythagorean theorem,$ON^2 + AN^2 = OA^2$.
So,$PN^2 - AN^2 = OP^2 - OA^2$.
Since $OA$ and $OT$ are both radii of the same circle,$OA = OT$.
Therefore,$PN^2 - AN^2 = OP^2 - OT^2$.
$(iii)$ From $(i)$ and $(ii)$,we have $PA \cdot PB = OP^2 - OT^2$.
In the right-angled triangle $\triangle OTP$ (where $\angle OTP = 90^{\circ}$ because the tangent is perpendicular to the radius at the point of contact),by the Pythagorean theorem:
$OP^2 = OT^2 + PT^2 \implies OP^2 - OT^2 = PT^2$.
Thus,$PA \cdot PB = PT^2$.

(N/A) As shown in the figure,the circle touches the sides of the triangle $ABC$ at points $P$,$Q$,and $R$.
According to the theorem,the lengths of tangents drawn from an external point to a circle are equal.
Therefore,we have:
$BQ = BP$
$CP = CR$
$AQ = AR$
Now,consider the perimeter of the triangle $ABC$:
$Perimeter = AB + BC + AC$
$Perimeter = AB + (BP + CP) + AC$
Substituting the values from the tangent theorem:
$Perimeter = AB + BQ + CR + AC$
$Perimeter = (AB + BQ) + (AC + CR)$
$Perimeter = AQ + AR$
Since $AQ = AR$,we can write:
$Perimeter = AQ + AQ = 2AQ$
Thus,$2AQ = AB + BC + AC$
$AQ = \frac{1}{2}(AB + BC + AC)$

(N/A) Let the points of contact of the circle with the sides $AB, BC, CD, DE, EF,$ and $FA$ be $P, Q, R, S, T,$ and $U$ respectively.
Since the lengths of tangents drawn from an external point to a circle are equal,we have:
$AP = AU$
$BP = BQ$
$CQ = CR$
$DR = DS$
$ES = ET$
$FT = FU$
Now,consider the sum of alternate sides:
$AB + CD + EF = (AP + PB) + (CR + RD) + (ET + TF)$
Substituting the tangent equalities:
$AB + CD + EF = (AU + BQ) + (CQ + DS) + (ES + FU)$
Rearranging the terms:
$AB + CD + EF = (BQ + CQ) + (DS + ES) + (FU + AU)$
$AB + CD + EF = BC + DE + FA$
Hence,it is proved.

(N/A) circle is inscribed in the $\triangle ABC$,which touches the sides $BC, CA$ and $AB$ at $D, E$ and $F$ respectively.
Given,$BC = a, CA = b$ and $AB = c$.
By using the property that tangents drawn from an external point to a circle are equal in length:
$\therefore BD = BF = x$ (let)
$DC = CE = y$ (let)
$AE = AF = z$ (let)
The perimeter of $\triangle ABC = BC + CA + AB = a + b + c$.
Since $2s = a + b + c$,we have:
$(BD + DC) + (CE + EA) + (AF + FB) = a + b + c$
$(x + y) + (y + z) + (z + x) = a + b + c$
$2(x + y + z) = 2s$
$x + y + z = s$
We need to find $BD = x$.
From the equation $x + y + z = s$,we get $x = s - (y + z)$.
Since $b = CA = CE + EA = y + z$,substituting this into the equation:
$x = s - b$
Therefore,$BD = s - b$.
Hence proved.

(B) Given that $PA$ and $PB$ are tangents from an external point $P$ to a circle with centre $O$.
We know that the lengths of tangents drawn from an external point to a circle are equal,so $PA = PB = 10 \, cm$.
The perimeter of $\triangle PCD = PC + CD + PD$.
Since $CD$ is a tangent at point $E$,it can be written as $CD = CE + ED$.
Also,$CE = CA$ and $ED = DB$ (tangents from points $C$ and $D$ respectively).
Substituting these into the perimeter formula:
Perimeter of $\triangle PCD = PC + (CE + ED) + PD = PC + CA + DB + PD$.
Since $PC + CA = PA$ and $PD + DB = PB$,we get:
Perimeter of $\triangle PCD = PA + PB = 10 + 10 = 20 \, cm$.

(N/A) Since $AC$ is a diameter,the angle in a semi-circle is $90^{\circ}$.
Therefore,$\angle ABC = 90^{\circ}$ (Angle in a semi-circle property).
In $\Delta ABC$,the sum of interior angles is $180^{\circ}$,so $\angle CAB + \angle ABC + \angle ACB = 180^{\circ}$.
Substituting $\angle ABC = 90^{\circ}$,we get $\angle CAB + \angle ACB = 180^{\circ} - 90^{\circ} = 90^{\circ}$ .......$(i)$
Since $AT$ is a tangent at $A$ and $AC$ is the radius (part of the diameter),the tangent is perpendicular to the radius at the point of contact.
Thus,$\angle CAT = 90^{\circ}$.
This implies $\angle CAB + \angle BAT = 90^{\circ}$ ..........$(ii)$
From equations $(i)$ and $(ii)$:
$\angle CAB + \angle ACB = \angle CAB + \angle BAT$
Subtracting $\angle CAB$ from both sides,we get $\angle ACB = \angle BAT$. Hence proved.

(D) Given,two circles have radii $OP = 3\, cm$ and $PO^{\prime} = 4\, cm$.
These two circles intersect at $P$ and $Q$.
Since $OP$ and $O^{\prime}P$ are radii to the points of contact of the tangents at $P$,and the problem states $OP$ and $O^{\prime}P$ are perpendicular to each other at $P$,we have $\angle OPO^{\prime} = 90^{\circ}$.
In right-angled $\triangle OPO^{\prime}$,by Pythagoras theorem:
$(OO^{\prime})^2 = (OP)^2 + (PO^{\prime})^2 = 3^2 + 4^2 = 9 + 16 = 25$.
So,$OO^{\prime} = 5\, cm$.
Let $PN$ be the perpendicular from $P$ to $OO^{\prime}$,where $N$ is the intersection point on $OO^{\prime}$.
Let $ON = x$,then $NO^{\prime} = 5 - x$.
In right-angled $\triangle OPN$,$(PN)^2 = (OP)^2 - (ON)^2 = 3^2 - x^2 = 9 - x^2$ ... $(i)$
In right-angled $\triangle PNO^{\prime}$,$(PN)^2 = (PO^{\prime})^2 - (NO^{\prime})^2 = 4^2 - (5 - x)^2 = 16 - (25 + x^2 - 10x) = 16 - 25 - x^2 + 10x = 10x - 9 - x^2$ ... (ii)
Equating $(i)$ and (ii):
$9 - x^2 = 10x - 9 - x^2$
$18 = 10x \Rightarrow x = 1.8\, cm$.
Substituting $x = 1.8$ in $(i)$:
$(PN)^2 = 9 - (1.8)^2 = 9 - 3.24 = 5.76$.
$PN = \sqrt{5.76} = 2.4\, cm$.
The common chord $PQ = 2 \times PN = 2 \times 2.4 = 4.8\, cm$.

(N/A) Let $O$ be the centre of the given circle. Suppose the tangent at $P$ meets $BC$ at $Q$. Join $BP$.
To prove: $BQ = QC$.
Proof: $\angle ABC = 90^{\circ}$.
In $\triangle ABC$,$\angle A + \angle C = 90^{\circ}$ (where $\angle A = \angle 1$ and $\angle C = \angle 5$).
So,$\angle 1 + \angle 5 = 90^{\circ}$ ... $(i)$
By the alternate segment theorem,the angle between the tangent at $P$ and the chord $BP$ is equal to the angle in the alternate segment,so $\angle 3 = \angle 1$.
Also,$\angle APB = 90^{\circ}$ (angle in a semi-circle).
Since $AC$ is a straight line,$\angle APB + \angle BPC = 180^{\circ}$,so $\angle BPC = 90^{\circ}$.
In $\triangle BPC$,$\angle 4 + \angle 5 = 90^{\circ}$ ... $(ii)$
From $(i)$ and $(ii)$:
$\angle 1 + \angle 5 = \angle 4 + \angle 5$
$\Rightarrow \angle 1 = \angle 4$
Since $\angle 3 = \angle 1$,we have $\angle 3 = \angle 4$.
In $\triangle PQC$,since $\angle 3 = \angle 4$,the sides opposite to equal angles are equal,so $PQ = QC$.
Also,tangents drawn from an external point to a circle are equal,so $PQ = BQ$.
Therefore,$BQ = QC$.
Hence,the tangent at $P$ bisects $BC$.

(B) $PQ$ and $PR$ are two tangents drawn from an external point $P$.
$\therefore PQ = PR$ [The lengths of tangents drawn from an external point to a circle are equal].
In $\triangle PQR$,since $PQ = PR$,the angles opposite to equal sides are equal.
$\therefore \angle PQR = \angle QRP$.
In $\triangle PQR$,the sum of angles is $180^{\circ}$.
$\angle PQR + \angle QRP + \angle RPQ = 180^{\circ}$
$2 \angle PQR + 30^{\circ} = 180^{\circ}$
$2 \angle PQR = 150^{\circ}$
$\angle PQR = 75^{\circ}$.
Since chord $SR \parallel$ tangent $PQ$,the alternate interior angles are equal.
$\therefore \angle SRQ = \angle RQP = 75^{\circ}$.
By the alternate segment theorem,the angle between the tangent $PR$ and chord $RQ$ is equal to the angle subtended by the chord in the alternate segment.
$\therefore \angle PQR = \angle QSR = 75^{\circ}$.
In $\triangle QRS$,the sum of angles is $180^{\circ}$.
$\angle RQS + \angle QSR + \angle SRQ = 180^{\circ}$
$\angle RQS + 75^{\circ} + 75^{\circ} = 180^{\circ}$
$\angle RQS + 150^{\circ} = 180^{\circ}$
$\angle RQS = 30^{\circ}$.

(N/A) $1$. In $\triangle ABC$,$\angle ACB = 90^{\circ}$ (Angle in a semicircle).
$2$. In $\triangle ABC$,$\angle ABC = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
$3$. Let $OC$ be the radius. Since $DC$ is a tangent,$OC \perp CD$. Thus,$\angle OCD = 90^{\circ}$.
$4$. $\angle OCA = \angle OAC = 30^{\circ}$ (Since $OA = OC$,$\triangle OAC$ is isosceles).
$5$. $\angle BCD = \angle OCD - \angle OCA = 90^{\circ} - 30^{\circ} = 60^{\circ}$.
$6$. In $\triangle BCD$,$\angle CBD = 180^{\circ} - \angle ABC = 180^{\circ} - 60^{\circ} = 120^{\circ}$ (Linear pair).
$7$. In $\triangle BCD$,$\angle BDC = 180^{\circ} - (120^{\circ} + 60^{\circ}) = 0^{\circ}$? Wait,let us re-evaluate: $\angle BCD = 60^{\circ}$,$\angle CBD = 120^{\circ}$,then $\angle BDC = 180^{\circ} - 180^{\circ} = 0^{\circ}$ is incorrect. Let's use the property: $\angle BCD = \angle BAC = 30^{\circ}$ (Angle between tangent and chord equals angle in alternate segment).
$8$. $\angle BCD = 30^{\circ}$. In $\triangle BCD$,$\angle CBD = 120^{\circ}$,$\angle BCD = 30^{\circ}$,so $\angle BDC = 180^{\circ} - (120^{\circ} + 30^{\circ}) = 30^{\circ}$.
$9$. Since $\angle BCD = \angle BDC = 30^{\circ}$,the triangle $BCD$ is isosceles with $BC = BD$.

(N/A) Let $M$ be the mid-point of an arc $AMB$ and $TMT'$ be the tangent to the circle at $M$.
Join $AB$,$AM$,and $MB$.
Since,$\text{arc } AM = \text{arc } MB$,
$\Rightarrow$ Chord $AM =$ Chord $MB$.
In $\triangle AMB$,$AM = MB$.
$\Rightarrow \angle MAB = \angle MBA$ (angles opposite to equal sides are equal) $\dots(i)$.
Since $TMT'$ is a tangent line at $M$,by the alternate segment theorem:
$\angle AMT = \angle MBA$ (angles in alternate segments are equal).
From $(i)$,$\angle AMT = \angle MAB$.
But $\angle AMT$ and $\angle MAB$ are alternate interior angles formed by the transversal $AM$ intersecting lines $AB$ and $TMT'$.
Since the alternate interior angles are equal,the lines must be parallel.
Therefore,$AB \parallel TMT'$.
Hence,the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

(N/A) Join $AO, OC$ and $O^{\prime}D, O^{\prime}B$.
In $\triangle E O^{\prime} D$ and $\triangle E O^{\prime} B$:
$O^{\prime} D = O^{\prime} B$ (radii of the same circle)
$O^{\prime} E = O^{\prime} E$ (common side)
$ED = EB$ (tangents drawn from an external point $E$ to the circle are equal in length)
Therefore,$\triangle E O^{\prime} D \cong \triangle E O^{\prime} B$ (by $SSS$ congruence rule).
This implies $\angle O^{\prime} E D = \angle O^{\prime} E B$,so $O^{\prime} E$ is the angle bisector of $\angle DEB$.
Similarly,by considering $\triangle EOA$ and $\triangle EOC$,we can prove $\triangle EOA \cong \triangle EOC$,which implies $\angle OEA = \angle OEC$,so $OE$ is the angle bisector of $\angle AEC$.
Since $AB$ and $CD$ are straight lines intersecting at $E$,$\angle AEC$ and $\angle DEB$ are vertically opposite angles,so $\angle AEC = \angle DEB$.
Since $OE$ bisects $\angle AEC$ and $O^{\prime}E$ bisects $\angle DEB$,and $\angle AEC = \angle DEB$,it follows that $\angle OEC = \angle O^{\prime}EB$.
Since $C, E, B$ lie on a straight line,$\angle OEC + \angle OEO^{\prime} + \angle O^{\prime}EB = 180^{\circ}$.
Substituting $\angle OEC = \angle O^{\prime}EB$,we get $2\angle O^{\prime}EB + \angle OEO^{\prime} = 180^{\circ}$.
However,since $O, E, O^{\prime}$ form a straight line,the angle $\angle OEO^{\prime}$ must be $180^{\circ}$.
Thus,$O, E, O^{\prime}$ are collinear.

(B) Given,$OT = 13 \, cm$ and radius $OP = 5 \, cm$.
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,$OP \perp PT$.
In right-angled $\triangle OPT$,by Pythagoras theorem:
$PT^2 = OT^2 - OP^2 = 13^2 - 5^2 = 169 - 25 = 144$.
So,$PT = 12 \, cm$.
Since the lengths of tangents drawn from an external point to a circle are equal,$PT = QT = 12 \, cm$.
Also,$ET = OT - OE = 13 - 5 = 8 \, cm$.
Let $PA = AE = x$. Then $AT = PT - PA = 12 - x$.
In right-angled $\triangle AET$,by Pythagoras theorem:
$AT^2 = AE^2 + ET^2$
$(12 - x)^2 = x^2 + 8^2$
$144 - 24x + x^2 = x^2 + 64$
$24x = 144 - 64 = 80$
$x = \frac{80}{24} = \frac{10}{3} \, cm$.
Similarly,$EB = QB = \frac{10}{3} \, cm$.
Therefore,$AB = AE + EB = \frac{10}{3} + \frac{10}{3} = \frac{20}{3} \, cm$.