Find the equation to the circle (x^{2} + y^{2} – 4x – 2y = 0) at the point (1, 3).
-
A.
2y – x -5 = 0 -
B.
2y + x – 5 = 0 -
C.
2y + x + 5 = 0 -
D.
2y – x + 5 = 0
Correct Answer: Option A
Explanation
We are given the equation (x^{2} + y^{2} – 4x – 2y = 0)
(y = x^{2} + y^{2} – 4x – 2y )
Using the method of implicit differentiation,
(frac{mathrm d y}{mathrm d x} = 2x + 2yfrac{mathrm d y}{mathrm d x} – 4 – 2frac{mathrm d y}{mathrm d x})
For the tangent, (frac{mathrm d y}{mathrm d x} = 0),
(therefore 2x + 2yfrac{mathrm d y}{mathrm d x} – 4 – 2frac{mathrm d y}{mathrm d x} = 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)