Ex 10.4, 1
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Given: Circle C1 with radii 5cm
& C2 with radii 3cm
Intersecting at P & Q.
OP = 5cm , XP = 3cm & OX = 4cm
To find: Length of common chord
i.e., length of PQ
Solution: Let the point where OX intersects PQ be R.
In Δ POX & Δ QOX
OP = OQ
XP = XQ
OX = OX
∴ Δ POX ≅ Δ QOX
∠ POX = ∠ QOX
Also,
In Δ POR & Δ QOR
OP = OQ
∠ POR = ∠ QOR
OR = OR
∴ Δ POR ≅ Δ QOR
⇒ ∠ PRO = ∠ QRO
& PR = RQ
Since PQ is a line
∠ PRO + ∠ QRO = 180°
∠ PRO + ∠ PRO = 180°
2∠ PRO = 180°
∠ PRO = (180°)/2
∠ PRO = 90°
Therefore,
∠ QRO = ∠ PRO = 90°
Also,
∠ PRX = ∠ QRO = 90°
Let OR = x,
So, XR = OX – OR = 4 – x
Now,
From (4) & (5)
52 – x2 = –7 – x2 + 8x
25 – x2 = –7 – x2 + 8x
25 + 7 – x2 + x2 = 8x
32 = 8x
8x = 32
x = 32/8
x = 4
Putting value of x in (4)
PR2 = 25 – x2
PR2 = 25 – 42
PR2 = 25 – 16
PR2 = 9
PR = √9 = 3
∴ PQ = 2PR = 2 × 3 = 6
Hence, length of common chord = 6 m
Note:
OR = x = 4 cm
& XR = 4 – x = 4 – 4 = 0 cm
since XR = 0
this means
point X & R coincide
Hence actual figure is as follows

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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