Question
Easy
The length of the common chord of the circles $(x-a)^{2}+y^{2}=c^{2}$ and $x^{2}+(y-b)^{2}=c^{2}$ is equal to :
1
$\sqrt{4c^{2}+a^{2}+b^{2}}$
2
$\sqrt{4c^{2}-a^{2}-b^{2}}$
3
$\sqrt{c^{2}+a^{2}+b^{2}}$
4
$\sqrt{c^{2}-a^{2}-b^{2}}$
Question Details
Time to Solve: 12
Exam: HTET
Level/Paper: Level 3
Chapter: Vectors and Coordinate Geometry
Topic: Two Dimensional Geometry
Correct Answer
Option B
Explanation
To determine the length of the common chord of the given circles, we need to analyze the geometric configuration of the circles and apply the appropriate mathematical principles. ### Explanation for Option 2 as the Correct Answer: The two circles given are: 1. Circle 1: \((x-a)^2 + y^2 = c^2\) 2. Circle 2: \(x^2 + (y-b)^2 = c^2\) Both circles have the same radius \(c\), but different centers: \((a, 0)\)…Read More
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