2y - x -5 = 0
2y + x - 5 = 0
2y + x + 5 = 0
2y - x + 5 = 0
Correct answer is A
We are given the equation x2+y2−4x−2y=0
y=x2+y2−4x−2y
Using the method of implicit differentiation,
dydx=2x+2ydydx−4−2dydx
For the tangent, dydx=0,
∴
(2y - 2)\frac{\mathrm d y}{\mathrm d x} = 4 - 2x \implies \frac{\mathrm d y}{\mathrm d x} = \frac{4 - 2x}{2y - 2}
At (1, 3), \frac{\mathrm d y}{\mathrm d x} = \frac{4 - 2(1)}{2(3) - 2} = \frac{2}{4} = \frac{1}{2}
Equation: \frac{y - 3}{x - 1} = \frac{1}{2} \implies 2y - 6 = x - 1
= 2y - x - 6 + 1 = 2y - x - 5 = 0
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