Question

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

• A. $x^2+y^2=9a^2$
• B. $x^2+y^2=16a^2$
• C. $x^2+y^2=4a^2$
• D. $x^2+y^2=a^2$

Answer 1

Answer: (C)

$x^2+y^2=4a^2$

Given, median of the equilateral triangle is $3a$ say $LD$
In $\Delta LMD$, we have

$(LM{)}^2=(LD{)}^2+(MD{)}^2$

$\Rightarrow (LM{)}^2=9a^2+{\left(\frac{LM}{2}\right)}^2$

$\Rightarrow \frac{3}{4}(LM{)}^2=9a^2$

$\Rightarrow (LM{)}^2=12a^2$

Again in triangle $OMD$,

$(OM{)}^2=(OD{)}^2+(MD{)}^2$

$\Rightarrow R^2=(3a-R{)}^2+{\left(\frac{LM}{2}\right)}^2$
$\Rightarrow R^2=9a^2+R^2-6aR+3a^2$

$\Rightarrow 6aR=12a^2$

$\Rightarrow R=2a$
So, equation of circle is

$(x-0{)}^2+(y-0{)}^2=(2a{)}^2$

$\Rightarrow x^2+y^2=4a^2$