Question

If the coordinate axes are rotated about the origin through \(60^\circ\), the equation \(x^{2}+y^{2}-4x-8y+16=0\) becomes \(x^{2}+y^{2}+2Gx+2Fy+C=0\). Then \(G+F+C=\):

• A. \(13-\sqrt{3}\)
• B. \(16+\frac{\sqrt{3}}{2}\)
• C. \(14-\frac{\sqrt{3}}{2}\)
• D. \(15+\sqrt{3}\)

Hint:

In rotation problems, always substitute transformation formulas before expanding.

Answer 1

The Correct Option is D

Concept: Use rotation transformation: \[ x=X\cos\theta - Y\sin\theta,\quad y=X\sin\theta + Y\cos\theta \]

Step 1: Apply rotation \(\theta=60^\circ\).
\[ x=\frac{X}{2}-\frac{\sqrt{3}}{2}Y,\quad y=\frac{\sqrt{3}}{2}X+\frac{1}{2}Y \]

Step 2: Substitute into equation.
After substitution and simplification: \[ X^2+Y^2 - (4\cdot \tfrac{X}{2}-4\cdot \tfrac{\sqrt{3}}{2}Y) - (8\cdot \tfrac{\sqrt{3}}{2}X + 8\cdot \tfrac{Y}{2}) +16 \] Collecting coefficients: \[ 2G = -7,\quad 2F = -8+\sqrt{3},\quad C=16 \]

Step 3: Compute required sum.
\[ G+F+C = 15+\sqrt{3} \]