Circles
NCERT Solutions • Class 9 Maths • Chapter 9Exercise 9.2
1. Two circles of radii $5 \text{ cm}$ and $3 \text{ cm}$ intersect at two points and the distance between their centres is $4 \text{ cm}$. Find the length of the common chord.
Let radius $r_1 = 5$, $r_2 = 3$ and distance $d = 4$.
Notice that $3^2 + 4^2 = 9 + 16 = 25 = 5^2$.
This satisfies the Pythagoras theorem. Thus, the center of the smaller circle lies on the common chord? No, it forms a right-angled triangle.
Let centres be $O$ and $O’$. Let the common chord be $AB$. $OO’ \perp AB$. Let $M$ be the intersection.
Let $OM = x$. Then $O’M = 4-x$.
In $\Delta OMA$: $5^2 – x^2 = AM^2$. In $\Delta O’MA$: $3^2 – (4-x)^2 = AM^2$.
$25 – x^2 = 9 – (16 + x^2 – 8x) \Rightarrow 25 = -7 + 8x \Rightarrow 32 = 8x \Rightarrow x = 4$.
Since $x=4$, $M$ coincides with $O’$.
Length of half chord $AM = \sqrt{5^2 – 4^2} = \sqrt{25-16} = \sqrt{9} = 3 \text{ cm}$.
Total length of common chord $= 2 \times 3 =$ $6 \text{ cm}$.
Notice that $3^2 + 4^2 = 9 + 16 = 25 = 5^2$.
This satisfies the Pythagoras theorem. Thus, the center of the smaller circle lies on the common chord? No, it forms a right-angled triangle.
Let centres be $O$ and $O’$. Let the common chord be $AB$. $OO’ \perp AB$. Let $M$ be the intersection.
Let $OM = x$. Then $O’M = 4-x$.
In $\Delta OMA$: $5^2 – x^2 = AM^2$. In $\Delta O’MA$: $3^2 – (4-x)^2 = AM^2$.
$25 – x^2 = 9 – (16 + x^2 – 8x) \Rightarrow 25 = -7 + 8x \Rightarrow 32 = 8x \Rightarrow x = 4$.
Since $x=4$, $M$ coincides with $O’$.
Length of half chord $AM = \sqrt{5^2 – 4^2} = \sqrt{25-16} = \sqrt{9} = 3 \text{ cm}$.
Total length of common chord $= 2 \times 3 =$ $6 \text{ cm}$.
2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Given: Chords $AB = CD$ intersect at $P$.
Construction: Draw perpendiculars $OM \perp AB$ and $ON \perp CD$. Join $OP$.
Proof:
In $\Delta OMP$ and $\Delta ONP$:
1. $OM = ON$ (Equal chords are equidistant from centre)
2. $\angle OMP = \angle ONP = 90^\circ$
3. $OP = OP$ (Common)
$\therefore \Delta OMP \cong \Delta ONP$ (RHS Rule) $\Rightarrow MP = NP$.
Since $AB = CD \Rightarrow AM = CN$ (Half of equal chords).
$AM + MP = CN + NP \Rightarrow$ $AP = CP$.
$AB – AP = CD – CP \Rightarrow$ $BP = DP$.
Construction: Draw perpendiculars $OM \perp AB$ and $ON \perp CD$. Join $OP$.
Proof:
In $\Delta OMP$ and $\Delta ONP$:
1. $OM = ON$ (Equal chords are equidistant from centre)
2. $\angle OMP = \angle ONP = 90^\circ$
3. $OP = OP$ (Common)
$\therefore \Delta OMP \cong \Delta ONP$ (RHS Rule) $\Rightarrow MP = NP$.
Since $AB = CD \Rightarrow AM = CN$ (Half of equal chords).
$AM + MP = CN + NP \Rightarrow$ $AP = CP$.
$AB – AP = CD – CP \Rightarrow$ $BP = DP$.
3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Given: Equal chords $AB$ and $CD$ intersect at $P$.
To Prove: $\angle OPE = \angle OPF$ (where E and F are feet of perpendiculars).
Proof:
From the congruence proved in Q2 ($\Delta OME \cong \Delta ONF$ or here $\Delta OMP \cong \Delta ONP$):
By CPCT, $\angle OPM = \angle OPN$.
Thus, the line $OP$ makes equal angles with the chords $AB$ and $CD$.
To Prove: $\angle OPE = \angle OPF$ (where E and F are feet of perpendiculars).
Proof:
From the congruence proved in Q2 ($\Delta OME \cong \Delta ONF$ or here $\Delta OMP \cong \Delta ONP$):
By CPCT, $\angle OPM = \angle OPN$.
Thus, the line $OP$ makes equal angles with the chords $AB$ and $CD$.
4. If a line intersects two concentric circles with centre $O$ at $A, B, C$ and $D$, prove that $AB = CD$.
Construction: Draw $OM \perp AD$.
Proof:
For the outer circle, $OM \perp AD$ bisects the chord $AD$.
$\therefore AM = DM$ … (1)
For the inner circle, $OM \perp BC$ bisects the chord $BC$.
$\therefore BM = CM$ … (2)
Subtracting (2) from (1):
$AM – BM = DM – CM$
$\Rightarrow$ $AB = CD$.
Proof:
For the outer circle, $OM \perp AD$ bisects the chord $AD$.
$\therefore AM = DM$ … (1)
For the inner circle, $OM \perp BC$ bisects the chord $BC$.
$\therefore BM = CM$ … (2)
Subtracting (2) from (1):
$AM – BM = DM – CM$
$\Rightarrow$ $AB = CD$.
5. Three girls Reshma, Salma and Mandip are standing on a circle of radius $5\text{m}$. Reshma to Salma = $6\text{m}$, Salma to Mandip = $6\text{m}$. What is the distance between Reshma and Mandip?
Let positions be $R, S, M$. Radius $OR=OS=OM=5$. $RS=SM=6$.
$ORSM$ is a kite since $OR=OM$ and $RS=SM$.
Diagonals intersect at $90^\circ$. Let $OS$ intersect $RM$ at $K$. $OS \perp RM$.
Let $KR = x$. In $\Delta ORK$ and $\Delta RSK$:
$RK^2 = 5^2 – OK^2$ and $RK^2 = 6^2 – (5-OK)^2$.
Let $OK = y$.
$25 – y^2 = 36 – (25 + y^2 – 10y) \Rightarrow 25 = 11 + 10y \Rightarrow 10y = 14 \Rightarrow y = 1.4$.
$RK^2 = 25 – (1.4)^2 = 25 – 1.96 = 23.04$.
$RK = \sqrt{23.04} = 4.8\text{m}$.
Total distance $RM = 2 \times 4.8 =$ $9.6\text{m}$.
$ORSM$ is a kite since $OR=OM$ and $RS=SM$.
Diagonals intersect at $90^\circ$. Let $OS$ intersect $RM$ at $K$. $OS \perp RM$.
Let $KR = x$. In $\Delta ORK$ and $\Delta RSK$:
$RK^2 = 5^2 – OK^2$ and $RK^2 = 6^2 – (5-OK)^2$.
Let $OK = y$.
$25 – y^2 = 36 – (25 + y^2 – 10y) \Rightarrow 25 = 11 + 10y \Rightarrow 10y = 14 \Rightarrow y = 1.4$.
$RK^2 = 25 – (1.4)^2 = 25 – 1.96 = 23.04$.
$RK = \sqrt{23.04} = 4.8\text{m}$.
Total distance $RM = 2 \times 4.8 =$ $9.6\text{m}$.
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6. A circular park of radius $20\text{m}$. Three boys Ankur, Syed and David are sitting at equal distance on its boundary. Find the length of the string of each phone.
Let positions be $A, S, D$. $\Delta ASD$ is equilateral. Let side be $x$.
The radius $R$ of the circumcircle of an equilateral triangle with side $x$ is given by:
$R = \frac{x}{\sqrt{3}}$.
Given $R = 20$.
$20 = \frac{x}{\sqrt{3}}$
$x = 20\sqrt{3}$.
Length of string = $20\sqrt{3}\text{ m}$.
The radius $R$ of the circumcircle of an equilateral triangle with side $x$ is given by:
$R = \frac{x}{\sqrt{3}}$.
Given $R = 20$.
$20 = \frac{x}{\sqrt{3}}$
$x = 20\sqrt{3}$.
Length of string = $20\sqrt{3}\text{ m}$.