Textbook - Circles Questions in English

(A) circle is a collection of infinite points on a plane that are at a constant distance from a fixed point called the center. Since a tangent to a circle touches the circle at exactly one point,and there are infinite points on the circumference of a circle,there can be infinite tangents to a circle.

(A) $(i)$ $A$ tangent to a circle touches the circle at exactly one point. Therefore,it intersects the circle in $1$ point.
$(ii)$ $A$ line that intersects a circle at two distinct points is known as a secant of the circle.

(A-D) $(i)$ $A$ circle can have at most $2$ parallel tangents. These tangents are parallel to each other and pass through the endpoints of a diameter.
$(ii)$ The common point of a tangent to a circle and the circle is called the point of contact.

(D) We know that the radius of a circle is perpendicular to the tangent at the point of contact.
Therefore,$OP \perp PQ$.
This forms a right-angled triangle $\triangle OPQ$,where $OP = 5 \, cm$ (radius) and $OQ = 12 \, cm$ (hypotenuse).
By applying the Pythagoras theorem in $\triangle OPQ$:
$OP^2 + PQ^2 = OQ^2$
$5^2 + PQ^2 = 12^2$
$25 + PQ^2 = 144$
$PQ^2 = 144 - 25$
$PQ^2 = 119$
$PQ = \sqrt{119} \, cm$.

(N/A) It can be observed from the figure that $AB$ and $CD$ are two parallel lines.
Line $AB$ intersects the circle at exactly two points,$P$ and $Q$. Therefore,line $AB$ is the secant of this circle.
Since line $CD$ intersects the circle at exactly one point,$R$,line $CD$ is the tangent to the circle.

(N/A) We are given two concentric circles $C_1$ and $C_2$ with centre $O$ and a chord $AB$ of the larger circle $C_1$ which touches the smaller circle $C_2$ at the point $P$. We need to prove that $AP = BP$.
Let us join $OP$. Then,$AB$ is a tangent to $C_2$ at $P$ and $OP$ is its radius.
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,we have:
$OP \perp AB$
Now,$AB$ is a chord of the circle $C_1$ and $OP \perp AB$. We know that the perpendicular drawn from the centre of a circle to a chord bisects the chord.
Therefore,$OP$ is the bisector of the chord $AB$,which implies:
$AP = BP$

(N/A) We are given a circle with centre $O$,an external point $T$,and two tangents $TP$ and $TQ$ to the circle,where $P$ and $Q$ are the points of contact (see figure).
We need to prove that $\angle PTQ = 2 \angle OPQ$.
Let $\angle PTQ = \theta$.
Now,$TP = TQ$ (tangents from an external point are equal in length). So,$\triangle TPQ$ is an isosceles triangle.
Therefore,$\angle TPQ = \angle TQP = \frac{1}{2} (180^{\circ} - \theta) = 90^{\circ} - \frac{1}{2} \theta$.
Also,the radius is perpendicular to the tangent at the point of contact,so $\angle OPT = 90^{\circ}$.
Thus,$\angle OPQ = \angle OPT - \angle TPQ = 90^{\circ} - (90^{\circ} - \frac{1}{2} \theta)$.
$\angle OPQ = \frac{1}{2} \theta = \frac{1}{2} \angle PTQ$.
This gives $\angle PTQ = 2 \angle OPQ$.

(D) Join $OT$. Let it intersect $PQ$ at the point $R$. Then $\triangle TPQ$ is isosceles and $TO$ is the angle bisector of $\angle PTQ$. So,$OT \perp PQ$ and therefore,$OT$ bisects $PQ$ which gives $PR = RQ = 4 \, cm$.
Also,$OR = \sqrt{OP^2 - PR^2} = \sqrt{5^2 - 4^2} \, cm = \sqrt{25 - 16} \, cm = \sqrt{9} \, cm = 3 \, cm$.
Now,$\angle TPR + \angle RPO = 90^{\circ}$ and $\angle TPR + \angle PTR = 90^{\circ}$ (since $\triangle TRP$ is a right triangle).
So,$\angle RPO = \angle PTR$.
Therefore,right triangle $TRP$ is similar to the right triangle $PRO$ by $AA$ similarity.
This gives $\frac{TP}{PO} = \frac{RP}{RO}$,i.e.,$\frac{TP}{5} = \frac{4}{3}$ or $TP = \frac{20}{3} \, cm$.

(A) Let $O$ be the centre of the circle.
Given that,
$OQ = 25 \, cm$ and $PQ = 24 \, cm$.
As the radius is perpendicular to the tangent at the point of contact,therefore $OP \perp PQ$.
Applying Pythagoras theorem in $\triangle OPQ$,we obtain:
$OP^2 + PQ^2 = OQ^2$
$OP^2 + 24^2 = 25^2$
$OP^2 + 576 = 625$
$OP^2 = 625 - 576$
$OP^2 = 49$
$OP = 7 \, cm$.
Therefore,the radius of the circle is $7 \, cm$.

(B) It is given that $TP$ and $TQ$ are tangents to the circle from an external point $T$.
The radius of a circle is perpendicular to the tangent at the point of contact.
Therefore,$OP \perp TP$ and $OQ \perp TQ$.
This implies $\angle OPT = 90^{\circ}$ and $\angle OQT = 90^{\circ}$.
In the quadrilateral $POQT$,the sum of all interior angles is $360^{\circ}$.
$\angle OPT + \angle POQ + \angle OQT + \angle PTQ = 360^{\circ}$
Substituting the known values:
$90^{\circ} + 110^{\circ} + 90^{\circ} + \angle PTQ = 360^{\circ}$
$290^{\circ} + \angle PTQ = 360^{\circ}$
$\angle PTQ = 360^{\circ} - 290^{\circ} = 70^{\circ}$

(C) Given that $PA$ and $PB$ are tangents to the circle from an external point $P$.
The radius drawn to the point of contact is perpendicular to the tangent.
Therefore,$OA \perp PA$ and $OB \perp PB$,which implies $\angle OAP = 90^{\circ}$ and $\angle OBP = 90^{\circ}$.
In the quadrilateral $OAPB$,the sum of interior angles is $360^{\circ}$.
$\angle OAP + \angle APB + \angle OBP + \angle AOB = 360^{\circ}$
$90^{\circ} + 80^{\circ} + 90^{\circ} + \angle AOB = 360^{\circ}$
$260^{\circ} + \angle AOB = 360^{\circ}$
$\angle AOB = 100^{\circ}$
In $\triangle OPA$ and $\triangle OPB$:
$OA = OB$ (Radii of the same circle)
$PA = PB$ (Tangents from an external point are equal)
$OP = OP$ (Common side)
By $SSS$ congruence criterion,$\triangle OPA \cong \triangle OPB$.
By $CPCT$,$\angle POA = \angle POB$.
Since $\angle POA + \angle POB = \angle AOB = 100^{\circ}$,we have $2 \angle POA = 100^{\circ}$.
Therefore,$\angle POA = 50^{\circ}$.

(N/A) Let $AB$ be a diameter of the circle with center $O$. Two tangents $PQ$ and $RS$ are drawn at points $A$ and $B$ respectively.
Since the radius is perpendicular to the tangent at the point of contact,we have:
$OA \perp RS$ and $OB \perp PQ$
Therefore:
$\angle OAR = 90^{\circ}$
$\angle OAS = 90^{\circ}$
$\angle OBP = 90^{\circ}$
$\angle OBQ = 90^{\circ}$
From the above,we can observe that:
$\angle OAR = \angle OBQ = 90^{\circ}$ (These are alternate interior angles)
$\angle OAS = \angle OBP = 90^{\circ}$ (These are alternate interior angles)
Since the alternate interior angles are equal,the lines $PQ$ and $RS$ must be parallel.

(N/A) Let us consider a circle with centre $O$. Let $AB$ be a tangent which touches the circle at $P$.
We have to prove that the line perpendicular to $AB$ at $P$ passes through the centre $O$. We shall prove this by the contradiction method.
Let us assume that the perpendicular to $AB$ at $P$ does not pass through the centre $O$. Let it pass through another point $O'$.
Join $OP$ and $O'P$.
As the perpendicular to $AB$ at $P$ passes through $O'$,therefore,
$\angle O'PB = 90^{\circ} \dots(1)$
$O$ is the centre of the circle and $P$ is the point of contact. We know that the radius through the point of contact is perpendicular to the tangent.
$\therefore \angle OPB = 90^{\circ} \dots(2)$
Comparing equations $(1)$ and $(2)$,we obtain
$\angle O'PB = \angle OPB \dots(3)$
From the figure,it can be observed that,
$\angle O'PB < \angle OPB \dots(4)$
Therefore,$\angle O'PB = \angle OPB$ is not possible.
It is only possible when the line $O'P$ coincides with $OP$.
Therefore,the perpendicular to $AB$ through $P$ passes through the centre $O$.

(B) Let the centre of the circle be $O$ and the point from which the tangent is drawn be $A$.
Let $B$ be the point of contact of the tangent on the circle.
Given that the distance of point $A$ from the centre $O$ is $OA = 5 \, cm$.
The length of the tangent $AB = 4 \, cm$.
We know that the radius of a circle is perpendicular to the tangent at the point of contact.
Therefore,$OB \perp AB$,which makes $\triangle ABO$ a right-angled triangle with $\angle OBA = 90^\circ$.
Applying the Pythagoras theorem in $\triangle ABO$:
$OA^2 = AB^2 + OB^2$
$5^2 = 4^2 + OB^2$
$25 = 16 + OB^2$
$OB^2 = 25 - 16$
$OB^2 = 9$
$OB = \sqrt{9} = 3 \, cm$.
Thus,the radius of the circle is $3 \, cm$.

(C) Let the two concentric circles be centered at point $O$. Let $PQ$ be the chord of the larger circle which touches the smaller circle at point $A$. Therefore,$PQ$ is tangent to the smaller circle at point $A$.
Since $OA$ is the radius of the smaller circle and $PQ$ is the tangent,$OA \perp PQ$.
Applying the Pythagoras theorem in $\triangle OAP$,we have:
$OA^2 + AP^2 = OP^2$
$3^2 + AP^2 = 5^2$
$9 + AP^2 = 25$
$AP^2 = 25 - 9 = 16$
$AP = \sqrt{16} = 4 \, cm$.
Since the perpendicular from the center of a circle to a chord bisects the chord,we have $AP = AQ$.
Therefore,the length of the chord $PQ = AP + AQ = 2 \times AP = 2 \times 4 = 8 \, cm$.

(N/A) We know that the lengths of tangents drawn from an external point to a circle are equal.
From point $D$,$DR = DS$ ... $(1)$
From point $C$,$CR = CQ$ ... $(2)$
From point $B$,$BP = BQ$ ... $(3)$
From point $A$,$AP = AS$ ... $(4)$
Adding equations $(1), (2), (3),$ and $(4)$,we get:
$DR + CR + BP + AP = DS + CQ + BQ + AS$
Rearranging the terms,we get:
$(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)$
Since $DR + CR = CD$,$BP + AP = AB$,$DS + AS = AD$,and $CQ + BQ = BC$,we have:
$CD + AB = AD + BC$
Hence,$AB + CD = AD + BC$.

(N/A) Join point $O$ to $C$.
In $\triangle OPA$ and $\triangle OCA$:
$OP = OC$ (Radii of the same circle)
$AP = AC$ (Tangents drawn from an external point $A$ are equal in length)
$AO = AO$ (Common side)
Therefore,$\triangle OPA \cong \triangle OCA$ by $SSS$ congruence criterion.
This implies $\angle POA = \angle COA$ ... $(i)$
Similarly,in $\triangle OQB$ and $\triangle OCB$:
$OQ = OC$ (Radii of the same circle)
$BQ = BC$ (Tangents drawn from an external point $B$ are equal in length)
$OB = OB$ (Common side)
Therefore,$\triangle OQB \cong \triangle OCB$ by $SSS$ congruence criterion.
This implies $\angle QOB = \angle COB$ ... $(ii)$
Since $POQ$ is a diameter of the circle,it is a straight line,so the sum of angles on one side is $180^{\circ}$:
$\angle POA + \angle COA + \angle COB + \angle QOB = 180^{\circ}$
Using equations $(i)$ and $(ii)$:
$2 \angle COA + 2 \angle COB = 180^{\circ}$
$2(\angle COA + \angle COB) = 180^{\circ}$
$\angle COA + \angle COB = 90^{\circ}$
Therefore,$\angle AOB = 90^{\circ}$.

(N/A) Let us consider a circle centered at point $O$. Let $P$ be an external point from which two tangents $PA$ and $PB$ are drawn to the circle,touching the circle at points $A$ and $B$ respectively. $AB$ is the line segment joining the points of contact $A$ and $B$,which subtends $\angle AOB$ at the center $O$ of the circle.
It can be observed that:
$OA$ (radius) $\perp PA$ (tangent)
Therefore,$\angle OAP = 90^{\circ}$
Similarly,$OB$ (radius) $\perp PB$ (tangent)
Therefore,$\angle OBP = 90^{\circ}$
In the quadrilateral $OAPB$,the sum of all interior angles is $360^{\circ}$.
$\angle OAP + \angle APB + \angle PBO + \angle BOA = 360^{\circ}$
$90^{\circ} + \angle APB + 90^{\circ} + \angle BOA = 360^{\circ}$
$\angle APB + \angle BOA = 180^{\circ}$
Hence,it is proved 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 center.

(N/A) Since $ABCD$ is a parallelogram,
$AB = CD \dots(1)$
$BC = AD \dots(2)$
It can be observed that:
$DR = DS$ (Tangents on the circle from point $D$)
$CR = CQ$ (Tangents on the circle from point $C$)
$BP = BQ$ (Tangents on the circle from point $B$)
$AP = AS$ (Tangents on the circle from point $A$)
Adding all these equations,we obtain:
$DR + CR + BP + AP = DS + CQ + BQ + AS$
$(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)$
$CD + AB = AD + BC$
On putting the values of equations $(1)$ and $(2)$ in this equation,we obtain:
$2AB = 2BC$
$AB = BC \dots(3)$
Comparing equations $(1), (2),$ and $(3),$ we obtain:
$AB = BC = CD = DA$
Hence,$ABCD$ is a rhombus.

(D) Let the circle touch the sides $AB$ and $AC$ of the triangle at points $E$ and $F$ respectively. Let the length of the tangent $AF$ be $x \, cm$.
Since tangents drawn from an external point to a circle are equal in length:
$CF = CD = 6 \, cm$
$BE = BD = 8 \, cm$
$AE = AF = x \, cm$
Thus,the sides of the triangle are:
$AB = AE + EB = x + 8$
$BC = BD + DC = 8 + 6 = 14 \, cm$
$AC = AF + FC = x + 6$
Semi-perimeter $s = \frac{(x+8) + 14 + (x+6)}{2} = \frac{2x + 28}{2} = x + 14$.
Using Heron's formula,Area of $\triangle ABC = \sqrt{s(s-a)(s-b)(s-c)}$
$= \sqrt{(x+14)(x+14-14)(x+14-(x+6))(x+14-(x+8))}$
$= \sqrt{(x+14)(x)(8)(6)} = \sqrt{48x(x+14)} = 4\sqrt{3x(x+14)}$.
Also,Area of $\triangle ABC = \text{Area}(\triangle OBC) + \text{Area}(\triangle OCA) + \text{Area}(\triangle OAB)$
$= \frac{1}{2} \times 4 \times 14 + \frac{1}{2} \times 4 \times (x+6) + \frac{1}{2} \times 4 \times (x+8)$
$= 28 + 2(x+6) + 2(x+8) = 28 + 2x + 12 + 2x + 16 = 56 + 4x = 4(14+x)$.
Equating the areas: $4\sqrt{3x(x+14)} = 4(14+x)$
$\sqrt{3x(x+14)} = 14+x$
Squaring both sides: $3x(x+14) = (14+x)^2$
$3x^2 + 42x = 196 + x^2 + 28x$
$2x^2 + 14x - 196 = 0 \Rightarrow x^2 + 7x - 98 = 0$
$(x+14)(x-7) = 0$. Since $x > 0$,$x = 7$.
Therefore,$AB = 7+8 = 15 \, cm$ and $AC = 7+6 = 13 \, cm$.

(N/A) Let $ABCD$ be a quadrilateral circumscribing a circle centered at $O$ such that it touches the circle at points $P, Q, R, S$. Let us join the vertices of the quadrilateral $ABCD$ to the center of the circle.
Consider $\triangle OAP$ and $\triangle OAS$:
$AP = AS$ (Tangents from the same point)
$OP = OS$ (Radii of the same circle)
$OA = OA$ (Common side)
Therefore,$\triangle OAP \cong \triangle OAS$ ($SSS$ congruence criterion).
Thus,$\angle POA = \angle AOS$,which implies $\angle 1 = \angle 8$.
Similarly,we can show that:
$\angle 2 = \angle 3$
$\angle 4 = \angle 5$
$\angle 6 = \angle 7$
Since the sum of angles around the center is $360^{\circ}$:
$\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^{\circ}$
Substituting the equal angles:
$(\angle 1 + \angle 8) + (\angle 2 + \angle 3) + (\angle 4 + \angle 5) + (\angle 6 + \angle 7) = 360^{\circ}$
$2\angle 1 + 2\angle 2 + 2\angle 5 + 2\angle 6 = 360^{\circ}$
$2(\angle 1 + \angle 2) + 2(\angle 5 + \angle 6) = 360^{\circ}$
$(\angle 1 + \angle 2) + (\angle 5 + \angle 6) = 180^{\circ}$
Thus,$\angle AOB + \angle COD = 180^{\circ}$.
Similarly,we can prove that $\angle BOC + \angle DOA = 180^{\circ}$.
Hence,opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.