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The length of the common chord of the tw...

The length of the common chord of the two circles of radii 10, 24 whose centres are 26 units apart is

A

`10/13` units

B

`240` units

C

`240/13` units

D

`120/13` units

Text Solution

AI Generated Solution

The correct Answer is:
To find the length of the common chord of two circles with radii 10 and 24 units, whose centers are 26 units apart, we can follow these steps: ### Step-by-Step Solution: 1. **Draw the Circles and Identify Points**: - Let the centers of the two circles be \( C_1 \) and \( C_2 \). - The radius of circle \( C_1 \) is \( r_1 = 10 \) and the radius of circle \( C_2 \) is \( r_2 = 24 \). - The distance between the centers \( C_1C_2 = d = 26 \). 2. **Label Points**: - Let \( A \) and \( B \) be the points where the common chord intersects the circles. - Let \( O \) be the midpoint of the common chord \( AB \). - Let \( OA = H \) (the distance from the center of circle \( C_1 \) to the midpoint of the chord). - Let \( OB = X \) (the distance from the center of circle \( C_2 \) to the midpoint of the chord). 3. **Apply Pythagorean Theorem**: - In triangle \( C_1AO \): \[ C_1A^2 = OA^2 + O C_1^2 \implies 10^2 = H^2 + X^2 \quad \text{(Equation 1)} \] - In triangle \( C_2BO \): \[ C_2B^2 = OB^2 + O C_2^2 \implies 24^2 = H^2 + (26 - X)^2 \quad \text{(Equation 2)} \] 4. **Substituting Values**: - From Equation 1: \[ 100 = H^2 + X^2 \quad \text{(1)} \] - From Equation 2: \[ 576 = H^2 + (26 - X)^2 \quad \text{(2)} \] 5. **Expand Equation 2**: - Expanding Equation 2: \[ 576 = H^2 + (676 - 52X + X^2) \] - Rearranging gives: \[ 576 = H^2 + 676 - 52X + X^2 \] - Thus: \[ 52X = 676 + H^2 - 576 + X^2 \] 6. **Subtract Equation 1 from Equation 2**: - Subtracting Equation 1 from the expanded Equation 2: \[ 576 - 100 = (H^2 + (26 - X)^2) - (H^2 + X^2) \] - This simplifies to: \[ 476 = 676 - 52X \] - Rearranging gives: \[ 52X = 200 \implies X = \frac{200}{52} = \frac{50}{13} \] 7. **Substituting Back to Find \( H \)**: - Substitute \( X \) back into Equation 1: \[ 100 = H^2 + \left(\frac{50}{13}\right)^2 \] - Calculate \( \left(\frac{50}{13}\right)^2 = \frac{2500}{169} \): \[ H^2 = 100 - \frac{2500}{169} = \frac{16900 - 2500}{169} = \frac{14400}{169} \] - Thus: \[ H = \frac{120}{13} \] 8. **Finding the Length of the Common Chord**: - The length of the common chord \( AB \) is given by: \[ AB = 2 \times OA = 2H = 2 \times \frac{120}{13} = \frac{240}{13} \] ### Final Answer: The length of the common chord is \( \frac{240}{13} \).
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