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(A) The vertices are $A(-a, 0)$ and $B(a, 0)$. The side length of the equilateral triangle is $2a$.Since the triangle is equilateral,the $x$-coordinat

Vedclass · Accepted Answer

$3x^2 + 3y^2 - 2\sqrt{3}ay = 3a^2$

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(A) The vertices are $A(-a, 0)$ and $B(a, 0)$. The side length of the equilateral triangle is $2a$.Since the triangle is equilateral,the $x$-coordinate of the third vertex $C$ is the midpoint of $AB$,which is $0$.The height of an equilateral triangle with side $s = 2a$ is $h = \frac{\sqrt{3}}{2} 	imes (2a) = \sqrt{3}a$.Since $C$ lies above the $x$-axis,its coordinates are $C(0, \sqrt{3}a)$.Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since $A(-a, 0)$ and $B(a, 0)$ lie on the circle:$a^2 - 2ga + c = 0$ and $a^2 + 2ga + c = 0$.Subtracting these gives $4ga = 0$,so $g = 0$.Then $a^2 + c = 0$,so $c = -a^2$.Since $C(0, \sqrt{3}a)$ lies on the circle:$0^2 + (\sqrt{3}a)^2 + 2f(\sqrt{3}a) - a^2 = 0$$3a^2 + 2\sqrt{3}af - a^2 = 0$$2a^2 + 2\sqrt{3}af = 0$$f = -\frac{a^2}{\sqrt{3}a} = -\frac{a}{\sqrt{3}}$.The equation is $x^2 + y^2 - \frac{2a}{\sqrt{3}}y - a^2 = 0$.Multiplying by $3$,we get $3x^2 + 3y^2 - 2\sqrt{3}ay - 3a^2 = 0$,or $3x^2 + 3y^2 - 2\sqrt{3}ay = 3a^2$.

If two vertices of an equilateral triangle are $A(-a, 0)$ and $B(a, 0)$,where $a > 0$,and the third vertex $C$ lies above the $x$-axis,then the equation of the circumcircle of $\Delta ABC$ is:

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