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JEE Main 2003Medium

Q.The equation of the circle with radius 3 and tangent to the line 3x - 4y + 5 = 0 is

a
x2+y2±6x±8y+25=0x^2 + y^2 \pm 6x \pm 8y + 25 = 0
b
x2+y2±6x±8y25=0x^2 + y^2 \pm 6x \pm 8y - 25 = 0
c
x2+y2±3x±4y+5=0x^2 + y^2 \pm 3x \pm 4y + 5 = 0
d
x2+y2±3x±4y5=0x^2 + y^2 \pm 3x \pm 4y - 5 = 0

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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