Two circles have the equations $x^{2} + y^{2} - 4 x - 6 y - 8 = 0$ and $x^{2} + y^{2} - 2 x - - 3 = 0$ . Then

they cut each other

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they touch each other

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one circle lies inside the other

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one circle lies wholly outside the other

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Solution

The correct option is A they cut each other
$C 1 (2, 3)$ and $r_{1} = \sqrt{2^{2} + 3^{2} + 8} = \sqrt{21}$
$C 2 (1, 0)$ and $r_{2} = \sqrt{1^{2} + 3} = 2$
$C 1 C 2 = \sqrt{1} 0$ and $r_{1} + r_{2} = \sqrt{2} 1 + 2$
Since, $C 1 C 2 < r_{1} + r_{2}$
Therefore, two circles intersect each other
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