<p>The tangent drawn from the origin to the circle $x^2 + y^2 + 2gx + 2fy + f^2 = 0$ are perpendicular, if:</p>
Step-by-Step Solution
Key Concept: Two tangents from an external point are perpendicular when the distance from the point to the centre equals $\sqrt{2}$ times the radius.
<p><strong>Analysis:</strong> The circle is $x^2 + y^2 + 2gx + 2fy + f^2 = 0$. Complete the square: $(x+g)^2 + (y+f)^2 = g^2 + f^2 - f^2 = g^2$. So the centre is $(-g, -f)$ and radius is $|g|$. For tangents from the origin $(0,0)$ to be perpendicular, the angle between them is $90°$. If two tangents from an external point to a circle are perpendicular, and the angle subtended at the centre is $\theta$, then $\theta = 90°$. For perpendicular tangents, if the distance from origin to centre is $d$ and radius is $r$, then $d^2 = 2r^2$. Here: $$g^2 + f^2 = 2g^2$$ $$f^2 = g^2$$ $$g = \pm f$$. Since the problem structure suggests $g = f$ (checking the options and standard form), we have $g = f$.</p><p>∴ Answer is (a).</p>
Correct Answer: a