Circle: The equation of a circle with origin as a center and passing through… | AIEEE 2002

AIEEE 2002 (Offline)Question 1

The equation of a circle with origin as a center and passing through an equilateral triangle whose median is of length $3a$ is :

Circle

$x^{2}+y^{2}=9a^{2}$$x^{2}+y^{2}=16a^{2}$$x^{2}+y^{2}=4a^{2}$✓ Correct answer$x^{2}+y^{2}=a^{2}$

Solution

Concept. The circle is the circumcircle; in an equilateral triangle the circumcentre is the centroid, which divides each median $2:1$. Formulas used. (1) $R=\frac23\times$ median.Step 1 - Use (1). $R=\frac23\cdot3a=2a$. Step 2 - Circle. $x^2+y^2=4a^2$. Answer: C

Circle - JEE Mathematics question (AIEEE 2002 (Offline))

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