If the line $a x + b y = c$ is a normal to the circle $x^{2} + y^{2} = 16$ , then

a = \frac{1}{2}, b = \frac{1}{2}

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c = 0

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a^{2} + b^{2} = 16

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\frac{a}{b} = 4

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Solution

The correct option is A $c = 0$
$x^{2} + y^{2} = 16$
Or
$(x - 0)^{2} + (y - 0)^{2} = 16$
Therefore the normal at any point of the circle should pass through the origin.
Hence
$a x + b y = c$ should pass through the origin.
This gives us $c = 0$ .

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If the line ax+by=c is a normal to the circle x2+y2=16 , then

The correct option is A c=0x2+y2=16Or (x−0)2+(y−0)2=16Therefore the normal at any point of the circle should pass through the origin.Hence ax+by=c should pass through the origin.This gives us c=0.

If the line $a x + b y = c$ is a normal to the circle $x^{2} + y^{2} = 16$ , then

The correct option is A $c = 0$
$x^{2} + y^{2} = 16$
Or
$(x - 0)^{2} + (y - 0)^{2} = 16$
Therefore the normal at any point of the circle should pass through the origin.
Hence
$a x + b y = c$ should pass through the origin.
This gives us $c = 0$ .