If two vertices of an equilateral triangle are `A (-a,0)and B (a,0), a gt 0,` and the third vertex C lies above y-axis, then the equation of the circumcircle of `Delta ABC` is :

Options

A. `3x ^(2) + 3y^(2) - 2 sqrt3 ay = 3a ^(2)`
B. `3x ^(2) + 3y ^(2) - 2ay = 3a ^(2)`
C. `x ^(2) +y ^(2) - 2ay =a ^(2)`
D. `x ^(2) +y ^(2) - sqrt3 ay =a ^(3)`

Answer

To find the equation of the circumcircle of the equilateral triangle with vertices A(-a, 0), B(a, 0), and C(0, b) where b is above the y-axis, we can follow these steps:

### Step 1: Identify the coordinates of the vertices
The coordinates of the vertices are:
- A = (-a, 0)
- B = (a, 0)
- C = (0, b)

### Step 2: Calculate the lengths of the sides of the triangle
Since ABC is an equilateral triangle, we need to find the lengths of the sides AC, BC, and AB.

- Length of AB:
\[ AB = \sqrt{(a - (-a))^2 + (0 - 0)^2} = \sqrt{(2a)^2} = 2a \]

- Length of AC:
\[ AC = \sqrt{(0 - (-a))^2 + (b - 0)^2} = \sqrt{a^2 + b^2} \]

- Length of BC:
\[ BC = \sqrt{(0 - a)^2 + (b - 0)^2} = \sqrt{a^2 + b^2} \]

### Step 3: Set the lengths equal
Since all sides of the equilateral triangle are equal, we set:
\[ AC = AB \]
\[ \sqrt{a^2 + b^2} = 2a \]

### Step 4: Square both sides to eliminate the square root
\[ a^2 + b^2 = (2a)^2 \]
\[ a^2 + b^2 = 4a^2 \]

### Step 5: Rearrange the equation to solve for b
\[ b^2 = 4a^2 - a^2 \]
\[ b^2 = 3a^2 \]
\[ b = \sqrt{3}a \]

### Step 6: Find the circumcenter of triangle ABC
The circumcenter (H) of an equilateral triangle is the same as its centroid. The coordinates of the centroid (H) are given by:
\[ H = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
Substituting the coordinates of A, B, and C:
\[ H = \left(\frac{-a + a + 0}{3}, \frac{0 + 0 + b}{3}\right) = \left(0, \frac{b}{3}\right) \]
Substituting \(b = \sqrt{3}a\):
\[ H = \left(0, \frac{\sqrt{3}a}{3}\right) \]

### Step 7: Calculate the radius of the circumcircle
The radius (R) is the distance from the circumcenter to any vertex, say A:
\[ R = \sqrt{(0 - (-a))^2 + \left(\frac{\sqrt{3}a}{3} - 0\right)^2} \]
\[ R = \sqrt{a^2 + \left(\frac{\sqrt{3}a}{3}\right)^2} = \sqrt{a^2 + \frac{3a^2}{9}} = \sqrt{a^2 + \frac{a^2}{3}} = \sqrt{\frac{4a^2}{3}} = \frac{2a}{\sqrt{3}} \]

### Step 8: Write the equation of the circumcircle
The standard equation of a circle with center (h, k) and radius r is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \(h = 0\), \(k = \frac{\sqrt{3}a}{3}\), and \(r = \frac{2a}{\sqrt{3}}\):
\[ (x - 0)^2 + \left(y - \frac{\sqrt{3}a}{3}\right)^2 = \left(\frac{2a}{\sqrt{3}}\right)^2 \]
\[ x^2 + \left(y - \frac{\sqrt{3}a}{3}\right)^2 = \frac{4a^2}{3} \]

### Step 9: Expand the equation
Expanding the left side:
\[ x^2 + \left(y^2 - 2y\frac{\sqrt{3}a}{3} + \frac{3a^2}{9}\right) = \frac{4a^2}{3} \]
\[ x^2 + y^2 - 2y\frac{\sqrt{3}a}{3} + \frac{a^2}{3} = \frac{4a^2}{3} \]

### Step 10: Rearranging the equation
Bringing all terms to one side gives:
\[ x^2 + y^2 - 2y\frac{\sqrt{3}a}{3} + \frac{a^2}{3} - \frac{4a^2}{3} = 0 \]
\[ x^2 + y^2 - 2y\frac{\sqrt{3}a}{3} - a^2 = 0 \]

### Final Equation
Multiplying through by 3 to eliminate fractions:
\[ 3x^2 + 3y^2 - 2\sqrt{3}ay - 3a^2 = 0 \]

### Conclusion
The equation of the circumcircle of triangle ABC is:
\[ 3x^2 + 3y^2 - 2\sqrt{3}ay - 3a^2 = 0 \]