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Equation of Tangent at a Point (x,y) in Terms of f'(x)

In the equati...

Question

In the equation $x^{2} + y^{2} - 4 x + 3 = 0$ is satisfied for real values of $x$ and $y$ , then

1 \leq x \leq 3

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2 \leq x \leq 3

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- \frac{1}{3} \leq y \leq \frac{1}{3}

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\frac{1}{3} \leq x \leq 1

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Solution

The correct option is C $1 \leq x \leq 3$
$x^{2} + y^{2} - 4 x + 3 = 0$
$∴ x^{2} - 4 x + 4 + y^{2} + 3 - 4 = 0$
$∴ {(x - 2)}^{2} + (y^{2} - 1) = 0$
$∴ x = \pm 2$ or $y = \pm 1$
$∴ 1 \leq x \leq 3$

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