Question

The equations of the tangents to the circle \( x^2 + y^2 = 36 \) which are perpendicular to the line \( 5x + y - 2 = 0 \) are

• A. \( x - 5y \pm 6\sqrt{26} = 0 \)
• B. \( x + 5y \pm 6\sqrt{26} = 0 \)
• C. \( x - 5y \pm \sqrt{26} = 0 \)
• D. \( x + 5y \pm \sqrt{26} = 0 \)

Hint:

Tangent to $x^2 + y^2 = r^2$ with slope $m$ is $y = mx \pm r\sqrt{1+m^2}$.

Answer 1

The Correct Option is A

Step 1: Find Slope
Given line: $5x + y - 2 = 0 \implies$ slope $m_1 = -5$.
Tangent is perpendicular, so its slope $m = \frac{-1}{m_1} = \frac{1}{5}$.
Step 2: Tangent Equation
For circle $x^2 + y^2 = r^2$, tangent is $y = mx \pm r\sqrt{1+m^2}$.
Here $r = 6$ and $m = 1/5$.
Step 3: Calculation
$y = \frac{1}{5}x \pm 6\sqrt{1 + (1/5)^2}$
$y = \frac{x}{5} \pm 6\sqrt{\frac{26}{25}} \implies 5y = x \pm 6\sqrt{26}$.
Step 4: Conclusion
$x - 5y \pm 6\sqrt{26} = 0$.
Final Answer:(A)