The tangent of circle $x^{2} + y^{2} - 2 x - 4 y + 1 = 0$ at $(1, 0)$ is

y = 0

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x = 0

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x + y = 0

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n o n e

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Solution

The correct option is A $y = 0$
The equation of circle is $x^{2} + y^{2} - 2 x - 4 x + 1 = 0$
The Point is $(1, 0)$
The slope of tangent is $2 x + 2 y \frac{d y}{d x} - 2 - 4 \frac{d y}{d x} = 0 (y - 2) \frac{d y}{d x} = 1 - x \frac{d y}{d x} = \frac{1 - x}{y - 2} {\frac{d y}{d x} ∣ ∣ ∣}_{(1, 0)} = 0$
Equation of tangent is $y - 0 = 0 (x - 1) y = 0$

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The tangent of circle x2+y2−2x−4y+1=0 at (1,0) is

The correct option is A y=0The equation of circle is x2+y2−2x−4x+1=0The Point is (1,0)The slope of tangent is 2x+2ydydx−2−4dydx=0(y−2)dydx=1−xdydx=1−xy−2dydx∣∣∣(1,0)=0Equation of tangent is y−0=0(x−1)y=0

The tangent of circle $x^{2} + y^{2} - 2 x - 4 y + 1 = 0$ at $(1, 0)$ is