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JEE Main 2010Medium

Q.The equation of the circle touching the line 2x + 3y + 1 = 0 at the point (1,-1) and passing through (4,1) is

a
x2+y2−6x−4y−12=0x^2 + y^2 - 6x - 4y - 12 = 0
b
x2+y2+6x+4y+12=0x^2 + y^2 + 6x + 4y + 12 = 0
c
x2+y2−6x+4y+12=0x^2 + y^2 - 6x + 4y + 12 = 0
d
x2+y2+6x−4y−12=0x^2 + y^2 + 6x - 4y - 12 = 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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The equation of the circle touching the line 2x + 3y + 1 = 0... | ParikshaNiti