Question : Two circles of radii 15 cm and 10 cm intersect each other and the length of their common chord is 16 cm. What is the distance (in cm) between their centres?
Option 1: $12+3 \sqrt{7}$
Option 2: $15+2 \sqrt{161}$
Option 3: $10+\sqrt{161}$
Option 4: $6+\sqrt{161}$
Correct Answer: $6+\sqrt{161}$
Solution :
Let PQ be a common chord. O and O' are the centres of the two circles such that
OP = 15 cm, O'P = 10 cm, and PQ = 16 cm
OO' perpendicularly bisects PQ at L.
So, PL = $\frac{1}{2}$ × PQ = $\frac{16}{2}$ = 8 cm
In $\triangle$ OLP, OP is hypotenuse
So, OP
2
= OL
2
+ PL
2
⇒ OL = $\sqrt{OP^{2} - PL^{2}}$
= $\sqrt{15^{2} - 8^{2}}$
= $\sqrt{161}$ cm
Similarly, in $\triangle$ O'LP, O'P is hypotenuse
So, O'P
2
= O'L
2
+ PL
2
⇒ O'L = $\sqrt{O'P^{2} - PL^{2}}$
= $\sqrt{10^{2} - 8^{2}}$
= $\sqrt{36}$
= 6 cm
So, OO' = O'L + OL
= 6 + $\sqrt{161}$
Hence, the correct answer is $6 + \sqrt{161}$.
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