The equation of a line parallel to the line $3 x + 4 y = 0$ and touching the circle $x^{2} + y^{2} = 9$ in the first quadrant is

3 x + 4 y = 15

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3 x + 4 y = 45

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3 x + 4 y = 9

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3 x + 4 y = 27

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Solution

The correct option is A $3 x + 4 y = 15$
Let a line parallel to the line $3 x + 4 y = 0$ be
$y = - \frac{3}{4} x + c \Rightarrow 4 y + 3 x - 4 c = 0$
Touches the circle $x^{2} + y^{2} = 9$
So distance from the centre $(0, 0)$ will be equal to radius,
$∣ ∣ ∣ ∣ \frac{- 4 c}{\sqrt{4^{2} + 3^{2}}} ∣ ∣ ∣ ∣ = 3 \Rightarrow 4 c = \pm 15$

We cannot take negative value as it will touch the circle in third quadrant so, required line is
$3 x + 4 y = 15$

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