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$S : x^{2} + y^{2} - 6 x + 4 y - 3 = 0$ is a circle and
$L : 4 x + 3 y + 19 = 0$ is a straight line.

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Correct option is D

:Centre of S is (3, – 2) and its radius is 32+(−2)2+3=4 Length of the perpendicular of (3, – 2) from the line L is 4(3)+3(−2)+19(4)2+(3)2=5 Which is greater yhen the radius ofS .So the line L lies outside the circle S

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S:x2+y2−6x+4y−3=0 is a circle andL:4x+3y+19=0 is a straight line.

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Ready to Test Your Skills?

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$S : x^{2} + y^{2} - 6 x + 4 y - 3 = 0$ is a circle and
$L : 4 x + 3 y + 19 = 0$ is a straight line.