Let the lengths of intercepts on x -axis and y -axis made by the circle x 2 + y 2 + a x + 2 a y + c = 0 , a…

Let the lengths of intercepts on x -axis and y -axis made by the circle x2+y2+ax+2ay+c=0, a<0 be 22 and 25, 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 :
  1. 11
  2. 7
  3. 6
  4. 10

Solution

x2+y2+ax+2ay+c=0

2g2-c=2a24-c=22

 a24-c=2 ...1

2f2-c=2a2-c=25

a2-c=5 ...2

1 & 2

3a24=3a=-2 a<0

  c=-1

Circle x2+y2-2x-4y-1=0

x-12+y-22=6

Given x+2y=0m=-12

mtangent=2

Equation of tangent y-2=2x-1±61+4

2x-y±30=0

Perpendicular distance from 0,0=±304+1=6

Asked in: JEE Main 2021 (16 Mar Shift 2)

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