Question

Suppose \(A\) and \(B\) are the points at which the line \[ x+y-\lambda=0 \] meets the pair of straight lines \[ x^2+y^2-2x-4y+2=0. \] If \(\angle AOB=90^\circ\), then the value of \(\lambda\) is:

• A. \(2\)
• B. \(3\)
• C. \(4\)
• D. \(0\)

Hint:

If two points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) subtend a right angle at the origin, then use \[ \overrightarrow{OA}\cdot\overrightarrow{OB}=x_1x_2+y_1y_2=0. \]

Answer 1

The Correct Option is A

Step 1: Use the line equation.
Given line is \[ x+y-\lambda=0. \] So, \[ y=\lambda-x. \]

Step 2: Substitute \(y=\lambda-x\) in the pair of lines equation.
Given, \[ x^2+y^2-2x-4y+2=0. \] Substituting \(y=\lambda-x\), \[ x^2+(\lambda-x)^2-2x-4(\lambda-x)+2=0. \] Expanding, \[ x^2+\lambda^2-2\lambda x+x^2-2x-4\lambda+4x+2=0. \] \[ 2x^2+(2-2\lambda)x+\lambda^2-4\lambda+2=0. \] Let the two intersection points be \[ A(x_1,\lambda-x_1),\quad B(x_2,\lambda-x_2). \]

Step 3: Use sum and product of roots.
From the quadratic equation, \[ 2x^2+(2-2\lambda)x+\lambda^2-4\lambda+2=0, \] we get \[ x_1+x_2=\lambda-1 \] and \[ x_1x_2=\frac{\lambda^2-4\lambda+2}{2}. \]

Step 4: Use the condition \(\angle AOB=90^\circ\).
Since \[ \angle AOB=90^\circ, \] we have \[ \overrightarrow{OA}\cdot \overrightarrow{OB}=0. \] Thus, \[ x_1x_2+(\lambda-x_1)(\lambda-x_2)=0. \] Expanding, \[ x_1x_2+\lambda^2-\lambda(x_1+x_2)+x_1x_2=0. \] \[ 2x_1x_2+\lambda^2-\lambda(x_1+x_2)=0. \]

Step 5: Substitute sum and product.
\[ 2\left(\frac{\lambda^2-4\lambda+2}{2}\right)+\lambda^2-\lambda(\lambda-1)=0. \] \[ \lambda^2-4\lambda+2+\lambda^2-\lambda^2+\lambda=0. \] \[ \lambda^2-3\lambda+2=0. \] Factorizing, \[ (\lambda-1)(\lambda-2)=0. \] So, \[ \lambda=1 \quad \text{or} \quad \lambda=2. \]

Step 6: Select the value from the options.
Among the given options, only \[ \lambda=2 \] is present.
Therefore, \[ \boxed{2} \]