Question:
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
Solution and Explanation
Concept:
The center of a circle passing through two points must lie on the perpendicular bisector of the segment connecting them.
Step 1: Find the perpendicular bisector of (0,0) and (4,0).
The points lie on the x-axis. The midpoint is \( (2, 0) \). The perpendicular bisector is the vertical line \( x = 2 \).
Step 2: Find the center.
The problem states the center lies on \( y = x \). Since the center must also be on \( x = 2 \), the center is \( (2, 2) \).
Step 3: Verify radius and form equation.
Radius \( r = \sqrt{8} \), so \( r^2 = 8 \). Center is \( (2, 2) \). Equation: \( (x - 2)^2 + (y - 2)^2 = 8 \).
Step 1: Find the perpendicular bisector of (0,0) and (4,0).
The points lie on the x-axis. The midpoint is \( (2, 0) \). The perpendicular bisector is the vertical line \( x = 2 \).
Step 2: Find the center.
The problem states the center lies on \( y = x \). Since the center must also be on \( x = 2 \), the center is \( (2, 2) \).
Step 3: Verify radius and form equation.
Radius \( r = \sqrt{8} \), so \( r^2 = 8 \). Center is \( (2, 2) \). Equation: \( (x - 2)^2 + (y - 2)^2 = 8 \).
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