Verified Solution • 2025–26 Session moderate

QJEEM21Ma7439: JEE Main 2021 Mathematics - Class 11 JEEM21Ma7439 Solution: Let the lengths of intercepts on x-axis and y-axis made by the…

Let the lengths of intercepts on x-axis and y-axis made by the circle
x2 + y2 + ax + 2ay + c = 0, (a < 0) be 2 2{\sqrt 2 } and 2 5{\sqrt 5 } , respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2y = 0, is equal to :

Multiple Choice Options

  • A 10{\sqrt {10} }
  • B 6{\sqrt {6} }
  • C 11{\sqrt {11} }
  • D 7{\sqrt {7} }

Step-by-Step Verified Solution

2a24−c=222\sqrt {{{{a^2}} \over 4} - c} = 2\sqrt 2

a2−4c=22\sqrt {{a^2} - 4c} = 2\sqrt 2

a2−4c=8{a^2} - 4c = 8 .... (1)

2a2−c=252\sqrt {{a^2} - c} = 2\sqrt 5

a2−c=5{a^2} - c = 5 .... (2)

(2)−(1)(2) - (1)

3c=−3a⇒c=−13c = - 3a \Rightarrow c = - 1

a2=4⇒a=−2{a^2} = 4 \Rightarrow a = - 2 (Given a < 0)

Equation of circle

x2+y2−2x−4y−1=0{x^2} + {y^2} - 2x - 4y - 1 = 0

Equation of tangent which is perpendicular to the line x + 2y = 0 is

2x−y+λ=02x - y + \lambda = 0

∴\therefore p = r

∣2−2+λ5∣=6\left| {{{2 - 2 + \lambda } \over {\sqrt 5 }}} \right| = \sqrt 6

⇒λ=±30\Rightarrow \lambda = \pm \sqrt {30}

∴\therefore Tangent 2x−y±30=02x - y \pm \sqrt {30} = 0

Distance from origin = 305=6{{\sqrt {30} } \over {\sqrt 5 }} = \sqrt 6

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring