KCET · Maths · Circle
The chord of the circle \(x^{2}+y^{2}-4 x=0\) which is bisected at \((1,0)\) is perpendicular to the line
- A \(y=x\)
- B \(x+y=0\)
- C \(\mathrm{x}=1\)
- D \(y=1\)
Answer & Solution
Correct Answer
(D) \(y=1\)
Step-by-step Solution
Detailed explanation
Given, equation of circle
\[
x^{2}+y^{2}-4 x=0 \text {, bisect by }(1,0)
\]
Centre of circle \(=(2,0)\)
Radius of circle \(=2\)
From figure, \(\mathrm{AB}\) is the chord of circle which bisected by the point \((1,0)\) and perpendicular to the line \(\mathrm{y}=1\).
Because it is parallel to x-axis while chord is parallel to \(\mathrm{y}\)-axis.
\[
x^{2}+y^{2}-4 x=0 \text {, bisect by }(1,0)
\]
Centre of circle \(=(2,0)\)
Radius of circle \(=2\)
From figure, \(\mathrm{AB}\) is the chord of circle which bisected by the point \((1,0)\) and perpendicular to the line \(\mathrm{y}=1\).
Because it is parallel to x-axis while chord is parallel to \(\mathrm{y}\)-axis.
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