Let the lengths of intercepts on x -axis and y -axis made by the circle x^2+y^2+ax+2ay+c=0, (a<0) be 2√2 and 2√5, respectively. Then the shortest…
JEE Main 2021 — Mathematics Coordinate Geometry
2021mcqmedium
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 :
Official previous-year question
Held on 16 Mar 2021 · Verified 6 Jul 2026.
Options
A
11
B
7
C
6
D
10
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Solution
x2+y2+ax+2ay+c=0
2g2−c=24a2−c=22
⇒4a2−c=2...(1)
2f2−c=2a2−c=25
⇒a2−c=5...(2)
(1)&(2)
43a2=3⇒a=−2(a<0)
∴c=−1
Circle ⇒x2+y2−2x−4y−1=0
⇒(x−1)2+(y−2)2=6
Given x+2y=0⇒m=−21
mtangent=2
Equation of tangent ⇒(y−2)=2(x−1)±61+4
⇒2x−y±30=0
Perpendicular distance from (0,0)=∣4+1±30∣=6
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