Question:
Find the equation of chord of the circle x²+y²=8x bisected at the point (4,3).
Find the equation of chord of the circle x²+y²=8x bisected at the point (4,3).
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Chord bisector passes through midpoint perpendicular to radius.
Updated On: Mar 18, 2026
- y=3
- y=1
- y=6
- y=7
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Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Center of circle is (4,0).
Step 2: Required chord is perpendicular to radius.
Step 3: Equation is y=3.
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