Two circles of radii $5 c m$ and $6 c m$ with common centre are drawn. There is a line $A B$ such that it is chord to both the circles. $C D = 8 c m .$ Find the distance of the chord from centre and the length of $A C$ respectively.

$3 c m, 1.19 c m$

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$2.55 c m, 2.35 c m$

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$3 c m, 2 c m$

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$2.16 c m, 2.35 c m$

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Solution

The correct option is A
$3 c m, 1.19 c m$

Given, $C D = 8 c m$

Since, $O E ⊥ C D$

$\Rightarrow C E = D E = \frac{C D}{2} = 4 c m$ .

Applying Pythagoras Theorem in $Δ O E C$ .

${O E}^{2} + {C E}^{2} = {O C}^{2}$

$\Rightarrow {O E}^{2} + 4^{2} = 5^{2}$

$∴ {O E}^{2} = 25 - 16$

$\Rightarrow O E = 3 c m$

In $Δ O E A$ ,

${O E}^{2} + {A E}^{2} = {O A}^{2}$

$\Rightarrow 3^{2} + {A E}^{2} = 6^{2}$

$\Rightarrow A E^{2} = 36 - 9 = 27$

$\Rightarrow A E = \sqrt{27} = 3 \sqrt{3} = 3 \times 1.73 = 5.19 c m$

Now, $A C = A E - C E = 5.19 - 4 = 1.19 c m$

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Two circles of radii 5 cm and 6 cm with common centre are drawn. There is a line AB such that it is chord to both the circles. CD=8 cm. Find the distance of the chord from centre and the length of AC respectively.

The correct option is A 3 cm, 1.19 cm Given, CD=8 cm Since, OE⊥CD ⇒CE=DE=CD2=4 cm. Applying Pythagoras Theorem in ΔOEC. OE2+CE2=OC2 ⇒OE2+42=52 ∴OE2=25−16 ⇒OE=3 cm In ΔOEA, OE2+AE2=OA2 ⇒32+AE2=62 ⇒AE2=36−9=27 ⇒AE=√27=3√3=3×1.73=5.19 cm Now, AC=AE−CE=5.19−4=1.19 cm

Two circles of radii $5 c m$ and $6 c m$ with common centre are drawn. There is a line $A B$ such that it is chord to both the circles. $C D = 8 c m .$ Find the distance of the chord from centre and the length of $A C$ respectively.