Two circle of radii 5 cm and 3 cm intersect two points and the distance between their centers is 4cn find the length vof the common CHORD
Answers
Answer
- The length of the common chord is 6 cm
Explanation
Given
- Radius of one circle is 5 cm
- Radius of the other one is 3 cm
- The distance between their centers will be 4 cm
To Find
- Length of the common chord
Solution
- OA = 5 cm
- AO' = 3 cm
- OO' = 4 cm
- AB = 2AC
In ∆AOC
→ AO² = AC² + OC²
→ 5² = (4-x)² + OC²
→ 25 = 16+x² - 8x + OC²
→ OC = 9-x²+8x -eq(1)
In ∆ACO'
→ AO'² = AC²+O'C²
→ 3² = AC²+x² -eq(2)
Equating eq(1) & eq(2)
→ 9-x²+8x = 9-x²
Cancelling the like terms
→ 8x = 0
→ x = 0
Substituting this in eq(1)
→ AC = 3 cm
Length of the chord will be
→ AB = 2AC
→ AB = 3×2
→ AB = 6 cm
Let two circles with centres O and O’ intersect each other at points A and B. On joining A and B, AB is a common chord.
Let two circles with centres O and O’ intersect each other at points A and B. On joining A and B, AB is a common chord.Radius OA = 5 cm, Radius O’A = 3 cm,
Let two circles with centres O and O’ intersect each other at points A and B. On joining A and B, AB is a common chord.Radius OA = 5 cm, Radius O’A = 3 cm,Distance between their centers OO’ = 4 cm
Let two circles with centres O and O’ intersect each other at points A and B. On joining A and B, AB is a common chord.Radius OA = 5 cm, Radius O’A = 3 cm,Distance between their centers OO’ = 4 cmIn triangle AOO’,
52 = 42 + 32
52 = 42 + 32 25 = 16 + 9
52 = 42 + 32 25 = 16 + 9 25 = 25
52 = 42 + 32 25 = 16 + 9 25 = 25Hence AOO’ is a right triangle, right angled at O’.
52 = 42 + 32 25 = 16 + 9 25 = 25 Hence AOO’ is a right triangle, right angled at O’.Since, perpendicular drawn from the center of the circle bisects the chord.
52 = 42 + 32 25 = 16 + 9 25 = 25 Hence AOO’ is a right triangle, right angled at O’.Since, perpendicular drawn from the center of the circle bisects the chord.Hence O’ is the mid-point of the chord AB. Also O’ is the centre of the circle II.
Therefore length of chord AB = Diameter of circle II