Question

The length of the common chord of the two circles $(x-a)²+(y-b)²=c²$ and $(x-b)²+(y-a)²=c²$ is

• A. $\sqrt{4c^{2}+2(a-b)^{2}}$
• B. $\sqrt{4c^{2}-2(a-b)^{2}}$
• C. $\sqrt{4c^{2}-(a-b)^{2}}$
• D. $\sqrt{2c^{2}-2(a-b)^{2}}$

Hint:

The length of the common chord of the two circles $(x-a)=c$ and $(x-b)+(y-a)=c$ is

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The common chord is found by subtracting the equations of the two circles ($S_1 - S_2 = 0$).
Step 2: Evaluation
$(x-a)^2 - (x-b)^2 + (y-b)^2 - (y-a)^2 = 0$, which simplifies to the line $x - y = 0$.
Step 3: Final Calculation
The distance from centre $(a, b)$ to the chord $x-y=0$ is $M = \frac{|a-b|}{\sqrt{2}}$. Length of chord = $2\sqrt{r^2 - M^2} = 2\sqrt{c^2 - \frac{(a-b)^2}{2}}$.
Step 4: Conclusion
Length $= \sqrt{4c^2 - 2(a-b)^2}$.
Hence, the Answer is: (b)