Question

The value of $\lambda$ for which the curve bisects the circumference of the circle $x^2+y^2+4x-2y+4=0$, is

• A. 1/3
• B. 5/3
• C. 3/5
• D. 3

Answer 1

Answer: (B)

$\frac{5}{3}$

Answer 2

1. Solve the Differential Equation: The given differential equation is $\frac{dy}{dx}=\frac{2x+\lambda}{\lambda-2y}$. Rearranging gives $(\lambda-2y)dy = (2x+\lambda)dx$. This can be rewritten as $2(y\,dy + x\,dx) = \lambda(dy - dx)$. Integrating both sides yields $x^2 + y^2 = \lambda(y - x) + C$.

2. Apply the Origin Condition: The problem states that the family of curves passes through the origin $(0,0)$ for every real value of $\lambda$. Substituting $(0,0)$ into the integrated equation gives $0^2 + 0^2 = \lambda(0 - 0) + C$, which implies $C=0$. Thus, the family of curves is represented by the equation $x^2 + y^2 = \lambda(y - x)$.

3. Analyze the Circle: The given circle is $x^2+y^2+4x-2y+4=0$. Completing the square, we rewrite it in standard form: $(x^2+4x+4) + (y^2-2y+1) + 4 - 4 - 1 = 0$ $(x+2)^2 + (y-1)^2 = 1$. This is a circle with center $(h, k) = (-2, 1)$ and radius $r=1$.

4. Apply the Bisecting Condition: A curve bisects the circumference of a circle if it passes through the center of the circle. Therefore, the curve $x^2 + y^2 = \lambda(y - x)$ must pass through the center of the given circle, which is $(-2, 1)$.

5. Solve for $\lambda$: Substitute the coordinates of the center $(-2, 1)$ into the curve's equation: $(-2)^2 + (1)^2 = \lambda(1 - (-2))$ $4 + 1 = \lambda(1 + 2)$ $5 = 3\lambda$ $\lambda = \frac{5}{3}$