Correct Answer - Option 1 : 4 sq. units
Concept:
The general equation for the circle
\(\rm ax^2+ay^2+2gx+2fy+c =0\)
The radius of that circle = \(\rm \sqrt{\left({g\over a}\right)^2+\left({f\over a}\right)^2-\left({c\over a}\right)}\)and
Center = \(\rm \left(-{g\over a},-{f\over a}\right)\)
Area of the triangle:
Area of the triangle with vertices \(\left(x_1, y_1\right), \left(x_2, y_2\right)\mbox{ and }\left(x_3, y_3\right)\) is given by:
\(|A| = \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\\ \end{vmatrix}\)
Calculation:
Given the equation of circle are
x2 + y2 + 2x + 4y - 4 = 0 ----(1)
x2 + y2 + 6x + 4y + 9 = 0 -----(2)
x2 + y2 - 4x - 4y + 2 = 0 ----(3)
We know that, center of circle
\(\rm ax^2+ay^2+2gx+2fy+c =0\)
are \(\rm \left(-{g\over a},-{f\over a}\right)\)
Hence, the center of given circles are
A(-1, -2), B(-3, -2), C(2, 2)
Hence, the area of a triangle ABC is
Δ = \(\frac{1}{2} \begin{vmatrix} -1 & -2 & 1\\ -3 & -2 & 1\\ 2& 2 & 1\\ \end{vmatrix}\)
Δ = \(\frac{1}{2}\)[-1(-2 - 2) + 2(-3 - 2) + 1(-6 + 4)]
⇒ Δ = 4 sq. unit
