Two circles of radii 5cm and 3cm intersect at two points and the distance between their centres is 4cm. Find the length of the common chord.
Answers
The distance between their centres is 4cm, the length of the common chord is 6 cm.
Step-by-step explanation:
Step 1:
The radius of the two circles is given as 5 cm and 3 cm respectively whose centres are O and O'.
∴ OA = OB = 5 cm
∴ O'A = O'B = 3 cm
If two circles intersect each other at two points, then the line joining their centres is the perpendicular bisector of the common cord.
So, OO' will be the perpendicular bisector of chord AB
∴ AC = BC = ½ * AB ….. (i)
Step 2:
Also it is given that the distance between their centres i.e., OO’ = 4 cm
Let OC be “x” then O'C will be “4 – x”.
By using Pythagoras theorem in right-angled Δ OAC,
OA² = OC² + AC²
⇒ 5² = x² + AC²
⇒ AC² = 25 − x² ….. (ii)
By using Pythagoras theorem in right-angled Δ O’AC,
O'A² = AC² + O'C²
⇒ 3² = AC² + (4 – x)²
⇒ 9 = AC² + 16 + x² − 8x
⇒ AC² = 8x − x² − 7 …… (iii)
Step 3:
Now, from (ii) and (iii), we get
25 − x² = 8x − x² − 7
⇒ 8x = 32
⇒ x = 4 cm
So, we get O’C = (4 -x) = 0 i.e., the common chord passes through the centre of the smaller circle, O' and thus, it will be the diameter of the smaller circle.
Step 4:
Substituting the value of x in eq. (ii), we get
AC² = 25 − x²
⇒ AC² = 25 − 4²
⇒ AC² = 25 − 16
⇒ AC² = 9
⇒ AC = 3 cm …. (iv)
Thus, from (i) & (iv), we get
The length of the common chord AB as,
= 2AC
= 2 * 3
= 6 cm
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