The normal line to a given curve at each point $(x, y)$ on the curve passes through the point $(3, 0)$ . If the curve contains the point $(3, 4)$ , then the equation of the curve is

x^{2} + y^{2} = 25

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x^{2} + y^{2} + 6 x = 43

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x^{2} + y^{2} - 6 x = 7

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x^{2} - y 2 + 6 x = 11

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Solution

The correct option is C $x^{2} + y^{2} - 6 x = 7$
Slope of the normal to any curve is $- \frac{d x}{d y}$
Now, equation of the normal which passes through $(3, 0)$ is
$y - 0 = - \frac{d x}{d y} (x - 3)$
$\Rightarrow y d y = - (x - 3) d x$
Integrating both sides, we get
$(x - 3)^{2} + y^{2} = C$
The curve passes through $(3, 4)$ .
$∴ C = 16$
So, the curve is $(x - 3)^{2} + y^{2} = 16$

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The normal line to a given curve at each point (x,y) on the curve passes through the point (3,0). If the curve contains the point (3,4), then the equation of the curve is

The correct option is C x2+y2−6x=7Slope of the normal to any curve is −dxdy Now, equation of the normal which passes through (3,0) is y−0=−dxdy(x−3) ⇒ydy=−(x−3)dx Integrating both sides, we get (x−3)2+y2=C The curve passes through (3,4). ∴C=16 So, the curve is (x−3)2+y2=16

The normal line to a given curve at each point $(x, y)$ on the curve passes through the point $(3, 0)$ . If the curve contains the point $(3, 4)$ , then the equation of the curve is