Exercise 9.2 Practice
Perpendicular from Centre to Chord
Q1: Intersecting Circles
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Two circles of radii $13 \text{ cm}$ and $15 \text{ cm}$ intersect at two points and the length of the common chord is $24 \text{ cm}$. Find the distance between their centres.
Let the two circles have centres $O$ and $O'$ and intersect at $A$ and $B$.
The common chord is $AB = 24 \text{ cm}$.
The line joining centres $OO'$ is the perpendicular bisector of the common chord $AB$.
Let $OO'$ intersect $AB$ at $M$.
$\therefore AM = \frac{1}{2} AB = \frac{24}{2} = 12 \text{ cm}$.
The common chord is $AB = 24 \text{ cm}$.
The line joining centres $OO'$ is the perpendicular bisector of the common chord $AB$.
Let $OO'$ intersect $AB$ at $M$.
$\therefore AM = \frac{1}{2} AB = \frac{24}{2} = 12 \text{ cm}$.
In right $\triangle OMA$:
$OA = 13 \text{ cm}$ (Radius of first circle).
$OM^2 = OA^2 - AM^2 = 13^2 - 12^2 = 169 - 144 = 25$.
$OM = 5 \text{ cm}$.
$OA = 13 \text{ cm}$ (Radius of first circle).
$OM^2 = OA^2 - AM^2 = 13^2 - 12^2 = 169 - 144 = 25$.
$OM = 5 \text{ cm}$.
In right $\triangle O'MA$:
$O'A = 15 \text{ cm}$ (Radius of second circle).
$O'M^2 = O'A^2 - AM^2 = 15^2 - 12^2 = 225 - 144 = 81$.
$O'M = 9 \text{ cm}$.
$O'A = 15 \text{ cm}$ (Radius of second circle).
$O'M^2 = O'A^2 - AM^2 = 15^2 - 12^2 = 225 - 144 = 81$.
$O'M = 9 \text{ cm}$.
Distance between centres $OO' = OM + O'M = 5 + 9 = 14 \text{ cm}$.
Distance = $14 \text{ cm}$
Q2: Concentric Circles
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A line $l$ intersects two concentric circles (circles with the same centre) with centre $O$ at points $A, B, C$ and $D$. Prove that $AB = CD$.
Draw $OM \perp l$.
For the outer circle, $AD$ is a chord and $OM \perp AD$.
Perpendicular from centre bisects the chord.
$\therefore AM = MD$ ... (i)
Perpendicular from centre bisects the chord.
$\therefore AM = MD$ ... (i)
For the inner circle, $BC$ is a chord and $OM \perp BC$.
$\therefore BM = MC$ ... (ii)
$\therefore BM = MC$ ... (ii)
Subtracting (ii) from (i):
$AM - BM = MD - MC$
$\Rightarrow AB = CD$.
$AM - BM = MD - MC$
$\Rightarrow AB = CD$.
Proved.
Q3: Equal Chords Intersection
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If two equal chords $AB$ and $CD$ of a circle intersect within the circle at point $P$, prove that $AP = CP$ and $BP = DP$.
Construction: Draw $OM \perp AB$ and $ON \perp CD$. Join $OP$.
In $\triangle OMP$ and $\triangle ONP$:
1. $\angle OMP = \angle ONP = 90^\circ$ (Construction)
2. $OP = OP$ (Common Hypotenuse)
3. $OM = ON$ (Equal chords are equidistant from centre)
$\therefore \triangle OMP \cong \triangle ONP$ (RHS Rule).
$\Rightarrow MP = NP$ (CPCT).
1. $\angle OMP = \angle ONP = 90^\circ$ (Construction)
2. $OP = OP$ (Common Hypotenuse)
3. $OM = ON$ (Equal chords are equidistant from centre)
$\therefore \triangle OMP \cong \triangle ONP$ (RHS Rule).
$\Rightarrow MP = NP$ (CPCT).
Since $AB = CD$, half of chords are equal: $AM = CN$.
Adding MP and NP:
$AM + MP = CN + NP \Rightarrow AP = CP$.
Adding MP and NP:
$AM + MP = CN + NP \Rightarrow AP = CP$.
Since $AB = CD$ and $AP = CP$, subtracting gives:
$AB - AP = CD - CP \Rightarrow BP = DP$.
$AB - AP = CD - CP \Rightarrow BP = DP$.
Proved.
Q4: Radius Calculation
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A chord of length $30 \text{ cm}$ is drawn at a distance of $8 \text{ cm}$ from the centre of a circle. Find the radius of the circle.
Let $AB$ be the chord and $OM$ be the perpendicular distance from centre $O$.
$AB = 30 \text{ cm}$, $OM = 8 \text{ cm}$.
$AB = 30 \text{ cm}$, $OM = 8 \text{ cm}$.
Perpendicular from centre bisects the chord.
$AM = MB = \frac{30}{2} = 15 \text{ cm}$.
$AM = MB = \frac{30}{2} = 15 \text{ cm}$.
In right $\triangle OMB$:
$OB^2 = OM^2 + MB^2$ (Pythagoras Theorem)
$OB^2 = 8^2 + 15^2$
$OB^2 = 64 + 225 = 289$.
$OB^2 = OM^2 + MB^2$ (Pythagoras Theorem)
$OB^2 = 8^2 + 15^2$
$OB^2 = 64 + 225 = 289$.
$OB = \sqrt{289} = 17 \text{ cm}$.
Radius = $17 \text{ cm}$
Q5: Parallel Chords
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Two parallel chords of lengths $10 \text{ cm}$ and $24 \text{ cm}$ are drawn in a circle of radius $13 \text{ cm}$. Find the distance between the chords if they are on:
(i) The same side of the centre.
(ii) Opposite sides of the centre.
(i) The same side of the centre.
(ii) Opposite sides of the centre.
Radius $r = 13 \text{ cm}$.
Chord 1: $AB = 24 \text{ cm}$. Half length = $12 \text{ cm}$.
Distance from centre $d_1 = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5 \text{ cm}$.
Chord 1: $AB = 24 \text{ cm}$. Half length = $12 \text{ cm}$.
Distance from centre $d_1 = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5 \text{ cm}$.
Chord 2: $CD = 10 \text{ cm}$. Half length = $5 \text{ cm}$.
Distance from centre $d_2 = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ cm}$.
Distance from centre $d_2 = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ cm}$.
(i) Same side:
Distance = $d_2 - d_1 = 12 - 5 = 7 \text{ cm}$.
Distance = $d_2 - d_1 = 12 - 5 = 7 \text{ cm}$.
(ii) Opposite sides:
Distance = $d_2 + d_1 = 12 + 5 = 17 \text{ cm}$.
Distance = $d_2 + d_1 = 12 + 5 = 17 \text{ cm}$.
Same side: $7 \text{ cm}$, Opposite: $17 \text{ cm}$