Circles(s) touching $x -$ axis at a distance $3$ from the origin and having an intercept of length $2 \sqrt{7}$ on $y -$ axis is (are)

x^{2} + y^{2} - 6 x + 8 y + 9 = 0

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x^{2} + y^{2} - 6 x + 7 y + 9 = 0

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x^{2} + y^{2} - 6 x - 8 y + 9 = 0

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Both A and C

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Solution

The correct option is D Both A and C
REF. Image
$k = \sqrt{9 + 7} = \sqrt{16} = 4$
$∴ 0^{l e}$ is
$(x - 3)^{2} + (y - 4)^{2} = 16$
$= x^{2} + y^{2} - 6 x - 8 y + 9 = 0$ or
$(x + 3)^{2} + (y - 4)^{2} = 16$
$= x^{2} + y^{2} + 6 x - 8 y + 9 = 0$ or
$(x - 3)^{2} + (y + 4)^{2} = 16$
Options A & C

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If the circles $x^{2} + y^{2} = a$ and $x^{2} + y^{2} - 6 x - 8 y + 9 = 0$ , touch externally, then a=

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x^{2} + 4 x + (y - 3)^{2} = 0

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