Question

Let S = 0 be a circle whose exactly one member of the given family of lines is it's tangent (L₂ = 0). If L₁ = 0 is also its tangent & L₁ = 0 & L₂ = 0 are perpendicular then radius of circle S = 0, is $\frac{10}{\sqrt{\lambda_1}}$, then the value of λ₁ is-

• A. 13
• B. 10
• C. 7
• D. 5

Answer 1

Answer: (A)

13

Answer 2

The family of lines is given by $(2x + y - 3) + \lambda(x + 2y - 3) = 0$. This family of lines always passes through a fixed point A. To find A, we set the coefficients of $\lambda$ to zero: $2x + y - 3 = 0$ $x + 2y - 3 = 0$ Solving these equations, we find $x=1$ and $y=1$. Thus, the fixed point is A(1, 1).

We are given $L_1: 3x + 2y + 5 = 0$. $L_2$ is a member of the family of lines and is tangent to circle S. The equation of a line in the family is $(2 + \lambda)x + (1 + 2\lambda)y - 3(1 + \lambda) = 0$. We are also given that $L_1$ and $L_2$ are perpendicular. The slope of $L_1$ is $m_1 = -\frac{3}{2}$. The slope of $L_2$ is $m_2 = -\frac{2 + \lambda}{1 + 2\lambda}$. For perpendicular lines, $m_1 m_2 = -1$: $(-\frac{3}{2}) \times (-\frac{2 + \lambda}{1 + 2\lambda}) = -1$ $\frac{3(2 + \lambda)}{2(1 + 2\lambda)} = -1$ $6 + 3\lambda = -2(1 + 2\lambda)$ $6 + 3\lambda = -2 - 4\lambda$ $7\lambda = -8 \implies \lambda = -\frac{8}{7}$.

The distance from the fixed point A(1,1) to the line $L_1: 3x + 2y + 5 = 0$ is: $d(A, L_1) = \frac{|3(1) + 2(1) + 5|}{\sqrt{3^2 + 2^2}} = \frac{|3 + 2 + 5|}{\sqrt{9 + 4}} = \frac{10}{\sqrt{13}}$.

The radius of circle S is given as $R = \frac{10}{\sqrt{\lambda_1}}$. Since the circle S is tangent to $L_1$ and $L_2$, and $L_1$ is a line, the radius of the circle is related to the distance from its center to $L_1$. The problem statement strongly suggests that the radius of the circle is equal to the distance from the fixed point A to $L_1$. Therefore, $R = \frac{10}{\sqrt{13}}$. Comparing this with the given formula for the radius, $R = \frac{10}{\sqrt{\lambda_1}}$, we have: $\frac{10}{\sqrt{\lambda_1}} = \frac{10}{\sqrt{13}}$ $\sqrt{\lambda_1} = \sqrt{13}$ $\lambda_1 = 13$.