Question:
The equations of the tangents to the circle $x^2 + y^2 = 36$ which are perpendicular to the line $5x + y = 2$, are
The equations of the tangents to the circle $x^2 + y^2 = 36$ which are perpendicular to the line $5x + y = 2$, are
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Perpendicular slope $= -1/m$. Use condition of tangency $c^2 = r^2(1+m^2)$.
Updated On: May 7, 2026
- $x + 5y \pm 6\sqrt{26} = 0$
- $x - 5y \pm 6\sqrt{26} = 0$
- $5x - y \pm 6\sqrt{26} = 0$
- $5x + y \pm 6\sqrt{26} = 0$
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The Correct Option is B
Solution and Explanation
Step 1: Slope of Tangent
Line: $5x + y = 2 \implies m_1 = -5$.
Tangent is perpendicular, so $m = -1/m_1 = 1/5$.
Step 2: Tangent Equation
For circle $x^2 + y^2 = r^2$, tangent is $y = mx \pm r\sqrt{1+m^2}$.
$r = 6, m = 1/5$.
Step 3: Calculation
$y = \frac{1}{5}x \pm 6\sqrt{1 + (1/25)} \implies 5y = x \pm 6\sqrt{26}/5 \cdot 5$.
$x - 5y \pm 6\sqrt{26} = 0$.
Final Answer: (B)
Line: $5x + y = 2 \implies m_1 = -5$.
Tangent is perpendicular, so $m = -1/m_1 = 1/5$.
Step 2: Tangent Equation
For circle $x^2 + y^2 = r^2$, tangent is $y = mx \pm r\sqrt{1+m^2}$.
$r = 6, m = 1/5$.
Step 3: Calculation
$y = \frac{1}{5}x \pm 6\sqrt{1 + (1/25)} \implies 5y = x \pm 6\sqrt{26}/5 \cdot 5$.
$x - 5y \pm 6\sqrt{26} = 0$.
Final Answer: (B)
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