Find the equation to the circle \(x^{2} + y^{2} - 4x - 2y = 0\)
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: A. 2y - x -5 = 0
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,
\(\frac{\mathrm d y}{\mathrm d x} = 2x + 2y\frac{\mathrm d y}{\mathrm d x} - 4 - 2\frac{\mathrm d y}{\mathrm d x}\)
For the tangent, \(\frac{\mathrm d y}{\mathrm d x} = 0\),
\(\therefore 2x + 2y\frac{\mathrm d y}{\mathrm d x} - 4 - 2\frac{\mathrm d y}{\mathrm d x} = 0\)
\((2y - 2)\frac{\mathrm d y}{\mathrm d x} = 4 - 2x \implies \frac{\mathrm d y}{\mathrm d x} = \frac{4 - 2x}{2y - 2}\)
At (1, 3), \(\frac{\mathrm d y}{\mathrm d x} = \frac{4 - 2(1)}{2(3) - 2} = \frac{2}{4} = \frac{1}{2}\)
Equation: \(\frac{y - 3}{x - 1} = \frac{1}{2} \implies 2y - 6 = x - 1\)
= \(2y - x - 6 + 1 = 2y - x - 5 = 0\)