Question

The equations of the tangents to the circle $x^2 + y^2 = 36$ which are perpendicular to the line $5x + y = 2$, are

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

Hint:

Perpendicular slope $= -1/m$. Use condition of tangency $c^2 = r^2(1+m^2)$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Perpendicular lines have slopes $m_1 \cdot m_2 = -1$.
Tangents to a circle at distance $r$ from the center must satisfy the length of perpendicular condition.
Step 2: Key Formula or Approach:
Line perpendicular to $Ax+By+C=0$ is $Bx-Ay+k=0$.
Given line $5x+y=2$. Perpendicular line is $x-5y+k=0$.
Distance from center $(0,0)$ to tangent $Ax+By+k=0$ is $r = \frac{|k|}{\sqrt{A^2+B^2}}$.
Step 3: Detailed Explanation:
Radius $r = \sqrt{36} = 6$.
Line is $x - 5y + k = 0$.
$\frac{|k|}{\sqrt{1^2 + (-5)^2}} = 6 \implies \frac{|k|}{\sqrt{26}} = 6 \implies k = \pm 6\sqrt{26}$.
Step 4: Final Answer:
The equations are $x - 5y \pm 6\sqrt{26} = 0$.