Question

The equation of the tangent to the circle \( x^2 + y^2 = 9 \) making an angle \( 60^\circ \) with the x-axis is:

• A. \( \frac{1}{\sqrt{3}}x - y \pm 6 = 0 \)
• B. \( \sqrt{3}x - y \pm 6 = 0 \)
• C. \( \sqrt{3}x + y \pm 6 = 0 \)
• D. \( x + y \pm \frac{6}{\sqrt{3}} = 0 \)

Hint:

Tangent to Circle with Given Slope}
General form for tangents: \( y = mx \pm r\sqrt{1 + m^2} \)
Angle with x-axis ⇒ slope = \( \tan(\theta) \)
Convert slope equation to general form: bring all terms to one side

Answer 1

The Correct Option is B

Circle: \( x^2 + y^2 = 9 \Rightarrow \) center \( = (0,0) \), radius \( r = 3 \) We are to find the tangent(s) to the circle making \( 60^\circ \) with x-axis. Slope \( m = \tan(60^\circ) = \sqrt{3} \) Equation of tangent to a circle from center at origin with slope \( m \) is: \[ y = mx \pm r\sqrt{1 + m^2} \Rightarrow y = \sqrt{3}x \pm 3\sqrt{1 + 3} = \sqrt{3}x \pm 6 \Rightarrow \sqrt{3}x - y \pm 6 = 0 \]