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Question

If the line ax+by=0 touches the circle x2+y2+2x+4y=0 and is a normal to the circle x2+y2−4x+2y−3=0, then value of (a,b) is/are

A
(1,2)
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B
(1,2)
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C
(1,2)
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D
(1,2)
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Solution

The correct options are
A (1,2)
D (1,2)
As the line ax+by=0 touches the circle x2+y2+2x+4y=0, so the distance of the centre (1,2) from the line is equal to radius
a2ba2+b2=(1)2+(2)20
Squaring both the sides, we get
(a+2b)2=5(a2+b2)
4a24ab+b2=0
(2ab)2=0
b=2a

Now, ax+by=0 is a normal to the circle x2+y24x+2y3=0, so it should pass through the centre (2,1)
2ab=0
b=2a
Only, (a,b)=(1,2) and (1,2) are satisfying the condition.

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