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JEE Main 202116 MarchEvening Shift
Exam
JEE Main
Year
2021
Date
16 March
Shift
Evening Shift
Subject
Mathematics
Chapter
Circle

JEE Main 2021 · 16 March · Evening Shift · Mathematics · Circle

Let the lengths of intercepts on x -axis and y -axis made by the...

Let the lengths of intercepts on xx -axis and yy -axis made by the circle x2+y2+ax+2ay+c=0,(a<0)be{x}^{2} + {y}^{2} + ax + 2ay + c = 0, \left(a < 0\right) be2\sqrt{2}andand2\sqrt{5},respectively.Thentheshortestdistancefromorigintoatangenttothiscirclewhichisperpendiculartotheline, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the linex + 2y = 0, is equal to :

Options

  1. A11\sqrt{11}
  2. B7\sqrt{7}
  3. C6\sqrt{6}
  4. D10\sqrt{10}

Correct answer

C. 6\sqrt{6}

Step-by-step solution

x2+y2+ax+2ay+c=0{x}^{2} + {y}^{2} + ax + 2ay + c = 0

2g2−c=2a24−c=222\sqrt{{g}^{2} - c} = 2\sqrt{\dfrac{{a}^{2}}{4} - c} = 2\sqrt{2}

⇒a24−c=2...(1)⇒ \dfrac{{a}^{2}}{4} - c = 2 . . . \left(1\right)

2f2−c=2a2−c=252\sqrt{{f}^{2} - c} = 2\sqrt{{a}^{2} - c} = 2\sqrt{5}

⇒a2−c=5...(2)⇒ {a}^{2} - c = 5 . . . \left(2\right)

\left(1\right) \text{ and } \left(2\right)</p><p></p><p>\dfrac{3{a}^{2}}{4} = 3 ⇒ a = - 2\left(a < 0\right)</p><p></p><p>∴ c = - 1</p><p>Circle</p><p>Circle⇒ {x}^{2} + {y}^{2} - 2x - 4y - 1 = 0</p><p></p><p>⇒ {\left(x - 1\right)}^{2} + {\left(y - 2\right)}^{2} = 6</p><p>Given</p><p>Givenx + 2y = 0 ⇒ m = - \dfrac{1}{2}</p><p></p><p>{m}_{tangent} = 2</p><p>Equationoftangent</p><p>Equation of tangent⇒ \left(y - 2\right) = 2\left(x - 1\right) \pm \sqrt{6}\sqrt{1 + 4}</p><p></p><p>⇒ 2x - y \pm \sqrt{30} = 0</p><p>Perpendiculardistancefrom(0,0)=∣±304+1∣=6</p><p>Perpendicular distance from \left(0, 0\right) = \left|\dfrac{\pm \sqrt{30}}{\sqrt{4 + 1}}\right| = \sqrt{6}

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