The figures in this solution are not accurate.
Section 1: Numerical Problems & Analytical Applications
Solution: Let $OD$ be the perpendicular distance from the centre $O$ to the chord $AB$.
Since the perpendicular from the centre bisects the chord: $$AD = \frac{1}{2} \times AB = \frac{1}{2} \times 8\text{ cm} = 4\text{ cm}$$ In right-angled triangle $\Delta OAD$, using Pythagoras Theorem: $$OA^2 = OD^2 + AD^2 \implies 5^2 = OD^2 + 4^2$$ $$25 = OD^2 + 16 \implies OD^2 = 25 - 16 = 9$$ $$OD = \sqrt{9} = 3\text{ cm}$$
Solution: Given that the Diameter $= 26\text{ cm}$. Therefore, Radius ($OP$) $= \frac{26}{2} = 13\text{ cm}$. Let $OD$ be the perpendicular from centre $O$ to chord $PQ$, so $OD = 5\text{ cm}$.
In right-angled triangle $\Delta OPD$, using Pythagoras Theorem: $$PD^2 = OP^2 - OD^2 = 13^2 - 5^2 = 169 - 25 = 144$$ $$PD = \sqrt{144} = 12\text{ cm}$$ Since $OD$ bisects the chord $PQ$: $$\text{Total Chord Length } PQ = 2 \times PD = 2 \times 12 = 24\text{ cm}$$
Solution: Let $OD \perp PQ$, so $OD = 2.1\text{ cm}$. The perpendicular bisects chord $PQ$, meaning $PD = \frac{4}{2} = 2\text{ cm}$.
In right-angled triangle $\Delta OPD$: $$PO^2 = OD^2 + PD^2 = (2.1)^2 + (2)^2 = 4.41 + 4 = 8.41$$ $$PO = \sqrt{8.41} = 2.9\text{ cm} \quad [\text{Radius}]$$ $$\text{Diameter} = 2 \times \text{Radius} = 2 \times 2.9 = 5.8\text{ cm}$$
Solution: Let the smaller chord be $AB = 6\text{ cm}$ and the larger chord be $CD = 8\text{ cm}$. Let $OP \perp AB$ and $OQ \perp CD$. Given $OP = 4\text{ cm}$.
Since the perpendicular bisects the chord, $AP = \frac{6}{2} = 3\text{ cm}$.
In $\Delta AOP$:
$$OA^2 = OP^2 + AP^2 = 4^2 + 3^2 = 16 + 9 = 25 \implies OA = 5\text{ cm} \quad [\text{Radius}]$$
Hence, $OC = 5\text{ cm}$ (also a radius). Now for chord $CD$, $CQ = \frac{8}{2} = 4\text{ cm}$.
In right-angled triangle $\Delta COQ$:
$$OQ^2 = OC^2 - CQ^2 = 5^2 - 4^2 = 25 - 16 = 9 \implies OQ = 3\text{ cm}$$
Solution: Let chord $AB = 48\text{ cm}$, with perpendicular distance $OE = 7\text{ cm}$. The midpoint $E$ gives $AE = 24\text{ cm}$.
In right-angled triangle $\Delta OAE$: $$OA^2 = AE^2 + OE^2 = 24^2 + 7^2 = 576 + 49 = 625 \implies OA = 25\text{ cm} \quad [\text{Radius}]$$ For the second chord $CD$, distance $OF = 20\text{ cm}$ and radius $OC = 25\text{ cm}$. In $\Delta COF$: $$CF^2 = OC^2 - OF^2 = 25^2 - 20^2 = 625 - 400 = 225 \implies CF = \sqrt{225} = 15\text{ cm}$$ $$\text{Total Length of second chord } CD = 2 \times CF = 2 \times 15 = 30\text{ cm}$$
Solution: Let the radius of the circle be $r\text{ cm}$, so $OA = OC = r$. $OP \perp AB$, which means $P$ is the midpoint of $AB$, so $AP = 3\text{ cm}$. Points $O, P, C$ lie on the same radial line.
Given $PC = 2\text{ cm}$, we can express $OP$ as: $$OP = OC - PC = (r - 2)\text{ cm}$$ In right-angled triangle $\Delta OAP$: $$OA^2 = OP^2 + AP^2 \implies r^2 = (r - 2)^2 + 3^2$$ $$r^2 = r^2 - 4r + 4 + 9 \implies r^2 = r^2 - 4r + 13$$ $$4r = 13 \implies r = \frac{13}{4} = 3.25\text{ cm}$$
Section 2: Geometric Proofs & Theorems
Given: Two concentric circles share a common center $O$. A straight line intersects the outer circle at $A$ and $B$, and the inner circle at $C$ and $D$.
To prove: $AC = BD$.
Construction: Draw a perpendicular line segment $OP$ from center $O$ to the straight line, so $OP \perp AB$.
Proof: For the outer circle, $AB$ is a chord and $OP \perp AB$. Since the perpendicular from the centre bisects a chord: $$AP = BP \quad \text{--- (Equation 1)}$$ For the inner circle, $CD$ is a chord and $OP \perp CD$. Therefore: $$CP = DP \quad \text{--- (Equation 2)}$$ Subtracting Equation 2 from Equation 1: $$AP - CP = BP - DP \implies AC = BD$$
Given: $AB$ and $CD$ are two chords of a circle with center $O$ that intersect at a point $P$, where $P$ is the midpoint of $AB$ ($AP = PB$).
To prove: $P$ cannot be the midpoint of $CD$ unless both chords pass through the center $O$ (i.e., they are diameters).
Proof: Since $P$ is the midpoint of chord $AB$, the line segment connecting the center to $P$ must be perpendicular to it:
$$\implies OP \perp AB \implies \angle OPB = 90^\circ$$
Now, if we assume $P$ is also the midpoint of chord $CD$, it would mean:
$$\implies OP \perp CD \implies \angle OPD = 90^\circ$$
This implies that $\angle OPB = \angle OPD = 90^\circ$ at the exact same point of intersection $P$. This is a geometric contradiction, as two distinct lines cannot be perpendicular to the same segment at the same point unless they coincide.
Therefore, $AB$ and $CD$ can only bisect each other if $P$ matches center $O$, which means both chords must be diameters.
Given: Two circles with centers $X$ and $Y$ intersect at points $A$ and $B$. $S$ is the midpoint of the line segment $XY$ ($XS = SY$). A line passing through $A$ is drawn perpendicular to $SA$ ($PQ \perp SA$), intersecting the circles at $P$ and $Q$.
To prove: $PA = AQ$.
Construction: Drop perpendiculars $XD \perp PQ$ and $YE \perp PQ$ from the centers $X$ and $Y$.
Proof: Since $XD$, $SA$, and $YE$ are all perpendicular to the same line $PQ$, they are parallel to each other: $$\implies XD \parallel SA \parallel YE$$ In the trapezoid $XDEY$, the lines $XD$, $SA$, and $YE$ are parallel, and $S$ is the midpoint of $XY$. According to the intercept theorem, parallel lines cut equal intercepts on any transversal line. Since $XS = SY$, it follows that: $$DA = AE \quad \text{--- (Equation 1)}$$ Since the perpendicular from the center of a circle bisects a chord, $XD \perp PA$ implies that $D$ is the midpoint of chord $PA$: $$DA = \frac{1}{2}PA \implies PA = 2 \cdot DA$$ Similarly, $YE \perp AQ$ implies that $E$ is the midpoint of chord $AQ$: $$AE = \frac{1}{2}AQ \implies AQ = 2 \cdot AE$$ Since $DA = AE$ from Equation 1, it follows that $2 \cdot DA = 2 \cdot AE$, which proves: $$PA = AQ$$
Given: Two circles with centers $P$ and $Q$ intersect at $A$ and $B$. A line through $A$ is parallel to $PQ$ ($CD \parallel PQ$) and intersects the circles at $C$ and $D$.
To prove: $CD = 2 \cdot PQ$.
Construction: Draw perpendiculars $PE \perp CD$ and $QF \perp CD$ from the centers $P$ and $Q$.
Proof: Since $PE \perp CD$ and $QF \perp CD$, the lines $PE$ and $QF$ are parallel to each other ($PE \parallel QF$). We are also given that $CD \parallel PQ$, which means $EF \parallel PQ$. Therefore, $PQFE$ forms a rectangle: $$\implies EF = PQ \quad \text{--- (Equation 1)}$$ The perpendicular from the center bisects a chord, so $PE \perp AC$ means $E$ is the midpoint of $AC$: $$AE = \frac{1}{2}AC$$ Similarly, $QF \perp AD$ means $F$ is the midpoint of $AD$: $$AF = \frac{1}{2}AD$$ The total length of the line segment $EF$ can be expressed as: $$EF = AE + AF = \frac{1}{2}AC + \frac{1}{2}AD = \frac{1}{2}(AC + AD) = \frac{1}{2}CD$$ Using the relation from Equation 1 ($EF = PQ$): $$PQ = \frac{1}{2}CD \implies CD = 2 \cdot PQ$$
Given: $AB$ and $AC$ are two equal chords ($AB = AC$) of a circle with center $O$.
To prove: The angle bisector of $\angle BAC$ passes through center $O$.
Construction: Join $OB$ and $OC$.
Proof: In triangles $\Delta AOB$ and $\Delta AOC$: $$AB = AC \quad [\text{Given}]$$ $$OB = OC \quad [\text{Radii of the same circle}]$$ $$AO = AO \quad [\text{Common side}]$$ Therefore, by Side-Side-Side (SSS) congruency: $$\Delta AOB \cong \Delta AOC \implies \angle OAB = \angle OAC \quad [\text{By C.P.C.T.}]$$ This shows that the line segment $AO$ bisects $\angle BAC$. Since this line segment connects vertex $A$ to the center $O$, the angle bisector passes directly through the center of the circle.
Given: $AB$ and $CD$ are two chords of a circle with center $O$. Their perpendicular distances from the center are $OP$ and $OQ$, respectively, where $OP < OQ$.
To prove: $AB > CD$.
Proof: Let $R$ be the radius of the circle ($OB = OD = R$). Since the perpendicular from the center bisects a chord, $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$: $$PB = \frac{1}{2}AB \quad \text{and} \quad QD = \frac{1}{2}CD$$ In right-angled triangle $\Delta OPB$, by Pythagoras Theorem: $$OB^2 = OP^2 + PB^2 \implies R^2 = OP^2 + \left(\frac{1}{2}AB\right)^2 \quad \text{--- (Equation 1)}$$ In right-angled triangle $\Delta OQD$, by Pythagoras Theorem: $$OD^2 = OQ^2 + QD^2 \implies R^2 = OQ^2 + \left(\frac{1}{2}CD\right)^2 \quad \text{--- (Equation 2)}$$ Equating Equation 1 and Equation 2: $$OP^2 + \left(\frac{1}{2}AB\right)^2 = OQ^2 + \left(\frac{1}{2}CD\right)^2$$ $$\left(\frac{1}{2}AB\right)^2 - \left(\frac{1}{2}CD\right)^2 = OQ^2 - OP^2$$ We are given that $OP < OQ \implies OQ^2 - OP^2 > 0$. Therefore: $$\left(\frac{1}{2}AB\right)^2 - \left(\frac{1}{2}CD\right)^2 > 0 \implies \left(\frac{1}{2}AB\right)^2 > \left(\frac{1}{2}CD\right)^2$$ $$\frac{1}{2}AB > \frac{1}{2}CD \implies AB > CD$$
Proof: Let $P$ be a given point inside a circle with center $O$. Let $AB$ be the chord passing through $P$ that is perpendicular to the diameter, so $OP \perp AB$. Let $CD$ be any other chord passing through $P$.
To show that $AB$ has the least length, we draw a perpendicular $OQ \perp CD$ from the center to the chord $CD$.
In right-angled triangle $\Delta OPQ$, the hypotenuse is always the longest side:
$$\implies OP > OQ$$
This means the perpendicular distance of chord $AB$ from the center ($OP$) is greater than the perpendicular distance of chord $CD$ from the center ($OQ$).
As proved in Question 12, a chord that is farther from the center is shorter than a chord that is closer to it. Since $OP > OQ$, it follows that:
$$AB < CD$$
Thus, $AB$ is the shortest possible chord passing through point $P$.
Section 3: Multiple Choice Questions (MCQs)
1. The lengths of two chords of a circle with centre O are equal. If $\angle AOB = 60^\circ$, then the value of $\angle COD$ is:
Chords of equal length subtend equal angles at the center of a circle. Since $AB = CD$, $\angle COD = \angle AOB = 60^\circ$.
Correct Option: (3) 60°
2. The length of the radius of a circle is $13\text{ cm}$ and the length of a chord is $10\text{ cm}$. The distance of the chord from the centre of the circle is:
The perpendicular from the center bisects the chord, so the base of the right-angled triangle is $\frac{10}{2} = 5\text{ cm}$. Using Pythagoras Theorem:
$$\text{Distance} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}$$
Correct Option: (2) 12 cm
3. AB and CD are two equal chords of a circle with its centre O. If the distance of the chord AB from the point O is $4\text{ cm}$, then the distance of the chord CD from the centre O is:
Chords of equal length are equidistant from the center of a circle. Since $AB = CD$, the distance of chord $CD$ from the center is also $4\text{ cm}$.
Correct Option: (2) 4 cm
4. The length of each of two parallel chords is $16\text{ cm}$. If the radius of the circle is $10\text{ cm}$, the distance between the two chords is:
The distance of each chord from the center is given by:
$$d = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6\text{ cm}$$
Since the chords lie on opposite sides of the center to maximize parallel spacing, the total distance between them is:
$$\text{Total Distance} = 6 + 6 = 12\text{ cm}$$
Correct Option: (1) 12 cm
5. The centre of two concentric circles is O; a straight line intersects one circle at points A and B and the other circle at points C and D. If AC = $5\text{ cm}$, then the length of BD is:
As proved in Question 7, for any line intersecting two concentric circles, $AC = BD$. Therefore, $BD = 5\text{ cm}$.
Correct Option: (2) 5 cm
Section 4: Short Answer Questions
Solution: Let the centers of the two circles be $A$ and $B$, and let their common chord be $CD = 12\text{ cm}$. The line segment joining the centers $AB$ is the perpendicular bisector of the common chord $CD$, intersecting it at point $O$.
Therefore, $CO = \frac{12}{2} = 6\text{ cm}$. The radius is $AC = 10\text{ cm}$.
In right-angled triangle $\Delta AOC$, using Pythagoras Theorem:
$$AO = \sqrt{AC^2 - CO^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8\text{ cm}$$
Since the circles are equal, the total distance between their centers is twice the length of $AO$:
$$\text{Distance } AB = 2 \times AO = 2 \times 8 = 16\text{ cm}$$
Solution: Let $O$ be the center of the circle, so radius $OA = OB = OC = 5\text{ cm}$. The line segment $AO$ extended is the perpendicular bisector of chord $BC$, intersecting it at point $P$. Let $AP = x$ and $BP = y$, so $BC = 2y$.
Since the center $O$ lies outside triangle $ABC$, point $P$ lies on the line segment $OA$. Therefore, $OP = 5 - x$.
In right-angled triangle $\Delta APB$:
$$x^2 + y^2 = AB^2 \implies x^2 + y^2 = 6^2 = 36 \quad \text{--- (Equation 1)}$$
In right-angled triangle $\Delta OPB$:
$$OP^2 + y^2 = OB^2 \implies (5 - x)^2 + y^2 = 5^2 = 25 \quad \text{--- (Equation 2)}$$
Subtracting Equation 2 from Equation 1:
$$x^2 - (5 - x)^2 = 36 - 25 \implies x^2 - (25 - 10x + x^2) = 11$$
$$10x - 25 = 11 \implies 10x = 36 \implies x = 3.6\text{ cm}$$
Substituting $x = 3.6$ back into Equation 1:
$$(3.6)^2 + y^2 = 36 \implies 12.96 + y^2 = 36$$
$$y^2 = 36 - 12.96 = 23.04 \implies y = \sqrt{23.04} = 4.8\text{ cm}$$
$$\text{Total Chord Length } BC = 2y = 2 \times 4.8 = 9.6\text{ cm}$$
Solution: We are given that $AB = CD = 6\text{ cm}$ because the chords are equal. The angle subtended at the center is $\angle AOB = 60^\circ$.
In triangle $\Delta AOB$, the sides $OA$ and $OB$ are radii of the circle, so $OA = OB$. This means the base angles are equal: $$\angle OAB = \angle OBA$$ The sum of angles in a triangle is $180^\circ$: $$\angle AOB + \angle OAB + \angle OBA = 180^\circ \implies 60^\circ + 2 \cdot \angle OAB = 180^\circ$$ $$2 \cdot \angle OAB = 120^\circ \implies \angle OAB = 60^\circ$$ Since all three angles are $60^\circ$, $\Delta AOB$ is an equilateral triangle. Therefore, its sides are equal: $$\text{Radius } OA = OB = AB = 6\text{ cm}$$
Solution: As proved in Question 13, the chord with the minimum length passing through a point $P$ is the one that is perpendicular to $OP$. Let this shortest chord be $AB$, so $OP \perp AB$.
In right-angled triangle $\Delta OPB$, the radius is $OB = 5\text{ cm}$ and $OP = 3\text{ cm}$. Using Pythagoras Theorem: $$BP = \sqrt{OB^2 - OP^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\text{ cm}$$ Since the perpendicular from the center bisects the chord, the total length of the shortest chord $AB$ is: $$\text{Least Length } AB = 2 \times BP = 2 \times 4 = 8\text{ cm}$$