JEE Main 2013 · 22 April · Online · Mathematics · Properties of Triangles
If two vertices of an equilateral triangle are A(-a, 0) and B(a,...
If two vertices of an equilateral triangle are A(−a,0) and B(a,0),a>0, and the third vertex C lies above x-axis then the equation of the circumcircle of △ABC is :
Options
A3x2+3y2−23ay=3a2
B3x2+3y2−2ay=3a2
Cx2+y2−2ay=a2
Dx2+y2−3ay=a2
Correct answer
A. 3x2+3y2−23ay=3a2
Step-by-step solution
Let C=(x,y)
Now, CA2=CB2=AB2⇒(x+a)2+y2=(x−a)2+y2=(2a)2⇒x2+2ax+a2+y2=4a2 and x2−2ax+a2+y2=4a2From(i)and(ii),x=0andy=\pm \sqrt{3} aSincepoint\mathrm{C}(x, y)liesabovethex−axisanda \gt 0,hencey=\sqrt{3} a∴\mathrm{C}=(0,\sqrt{3}a)Lettheequationofcircumcirclebex2+y2+2gx+2fy+\mathrm{C}=0SincepointsA(-a, 0), \mathrm{B}(a, 0) and C(0,3a) lie on the circle, therefore
a2−2ga+C=0a2+2ga+C=0 and 3 a^2+2 \sqrt{3} a f+\mathrm{C}=0From(iii),(iv),and(v)g=0,c=−a2,f=−\frac{a}{\sqrt{3}}Henceequationofthecircumcircleisx2+y2−32ay−a2=0⇒x2+y2−323ay−a2=0⇒3x2+3y2−23ay=3a2$
