If area of an equilateral triangle inscribed in the circle `x^2+y^2+10x+12y+c=0` is `27sqrt3`, then the value of c is
(a) 25
(b) -25
(c) 36
(d) -36

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Rewrite the equation of the circle
The given equation of the circle is:
\[ x^2 + y^2 + 10x + 12y + c = 0 \]

We can complete the square for the \(x\) and \(y\) terms.

### Step 2: Completing the square
For \(x\):
\[ x^2 + 10x = (x + 5)^2 - 25 \]

For \(y\):
\[ y^2 + 12y = (y + 6)^2 - 36 \]

Substituting these back into the equation gives:
\[ (x + 5)^2 - 25 + (y + 6)^2 - 36 + c = 0 \]
This simplifies to:
\[ (x + 5)^2 + (y + 6)^2 + c - 61 = 0 \]
Thus, we can rewrite it as:
\[ (x + 5)^2 + (y + 6)^2 = 61 - c \]

### Step 3: Identify the radius
From the standard form of the circle \((x - h)^2 + (y - k)^2 = r^2\), we see that:
\[ r^2 = 61 - c \]

### Step 4: Area of the equilateral triangle
The area \(A\) of an equilateral triangle inscribed in a circle of radius \(r\) is given by:
\[ A = \frac{3\sqrt{3}}{4} r^2 \]
Given that the area is \(27\sqrt{3}\), we can set up the equation:
\[ \frac{3\sqrt{3}}{4} r^2 = 27\sqrt{3} \]

### Step 5: Solve for \(r^2\)
To eliminate \(\sqrt{3}\), we divide both sides by \(\sqrt{3}\):
\[ \frac{3}{4} r^2 = 27 \]
Multiplying both sides by \(4\):
\[ 3r^2 = 108 \]
Now divide by \(3\):
\[ r^2 = 36 \]

### Step 6: Substitute \(r^2\) back into the equation
We know from Step 3 that:
\[ r^2 = 61 - c \]
Substituting \(r^2 = 36\):
\[ 36 = 61 - c \]

### Step 7: Solve for \(c\)
Rearranging gives:
\[ c = 61 - 36 = 25 \]

### Conclusion
Thus, the value of \(c\) is:
\[ \boxed{25} \]