Question:medium
A circle of radius \( \sqrt{8} \) is passing through origin and the point \( (4, 0) \). If the centre lies on the line \( y = x \), then the equation of the circle is:
A circle of radius \( \sqrt{8} \) is passing through origin and the point \( (4, 0) \). If the centre lies on the line \( y = x \), then the equation of the circle is:
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Always use the perpendicular bisector of chords to locate the center. If a chord is on an axis, the bisector is simply a line parallel to the other axis.
Updated On: May 6, 2026
- \( (x - 2)^2 + (y - 2)^2 = 8 \)
- \( (x + 2)^2 + (y + 2)^2 = 8 \)
- \( (x - 3)^2 + (y - 3)^2 = 8 \)
- \( (x + 3)^2 + (y + 3)^2 = 8 \)
- \( (x - 4)^2 + (y - 4)^2 = 8 \)
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The Correct Option is A
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