Length of intercepts made by circle $x^{2} + y^{2} - 10 x - 8 y + 4 = 0$ on the $X$ and $Y$ axes respectively are

\sqrt{21}, 2 \sqrt{3}

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2 \sqrt{21}, 2 \sqrt{3}

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2 \sqrt{21}, 4 \sqrt{3}

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\sqrt{3}, 2 \sqrt{21}

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Solution

The correct option is C $2 \sqrt{21}, 4 \sqrt{3}$
Given equation: $x^{2} + y^{2} - 10 x - 8 y + 4 = 0$
Comparing with $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
$g = - 5, f = - 4, c = 4$
Now,
Intercept made on $X - axis$
$= 2 \sqrt{g^{2} - c} = 2 \sqrt{(- 5)^{2} - 4} = 2 \sqrt{21}$
Intercept made on $Y - axis$
$= 2 \sqrt{f^{2} - c} = 2 \sqrt{(- 4)^{2} - 4} = 4 \sqrt{3}$

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x^{2} + y^{2} - 10 x - 8 y + 4 = 0

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(a) (\frac{1}{\sqrt{21}}, \frac{2}{\sqrt{21}}, \frac{4}{\sqrt{21}}) (b) (- \frac{1}{\sqrt{21}}, - \frac{2}{\sqrt{21}}, - \frac{4}{\sqrt{21}})

(c) (- \frac{1}{\sqrt{21}}, - \frac{2}{\sqrt{21}}, \frac{4}{\sqrt{21}}) (d) (\frac{1}{\sqrt{21}}, \frac{2}{\sqrt{21}}, - \frac{4}{\sqrt{21}})

x^{2} + y^{2} - 10 x + 6 y + 9 = 0

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