Question:
Let a circle of radius 3 pass through the origin \( O \) and meet the coordinate axes at the points \( A \) and \( B \). Then the locus of the centroid of the triangle \( AOB \) is :
Let a circle of radius 3 pass through the origin \( O \) and meet the coordinate axes at the points \( A \) and \( B \). Then the locus of the centroid of the triangle \( AOB \) is :
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The angle AOB is 90 degrees, so AB is a diameter of length 6; use the centroid formula.
Updated On: Nov 12, 2026
- \( x^2 + y^2 = 1 \)
- \( x^2 + y^2 = 4 \)
- \( 3x^2 + 3y^2 = 4 \)
- \( x^2 + y^2 = 12 \)
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The Correct Option is B
Solution and Explanation
The axes meet at O at a right angle, so \( \angle AOB = 90^\circ \) is an angle in a semicircle and AB is a diameter of the circle: \( AB = 6 \).
Let \( A(a, 0) \) and \( B(0, b) \). Then \( a^2 + b^2 = AB^2 = 36 \).
The centroid of triangle AOB is \( (h, k) = \left(\frac{a}{3}, \frac{b}{3}\right) \), so \( a = 3h \), \( b = 3k \).
\( 9h^2 + 9k^2 = 36 \Rightarrow h^2 + k^2 = 4 \).
Locus: \( x^2 + y^2 = 4 \).
Answer: (2) \( x^2 + y^2 = 4 \)
Let \( A(a, 0) \) and \( B(0, b) \). Then \( a^2 + b^2 = AB^2 = 36 \).
The centroid of triangle AOB is \( (h, k) = \left(\frac{a}{3}, \frac{b}{3}\right) \), so \( a = 3h \), \( b = 3k \).
\( 9h^2 + 9k^2 = 36 \Rightarrow h^2 + k^2 = 4 \).
Locus: \( x^2 + y^2 = 4 \).
Answer: (2) \( x^2 + y^2 = 4 \)
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