Question

For all real values of \(\lambda\), the point that lies on the polar of \[ (2\lambda,\lambda-4) \] with respect to the circle \[ x^2+y^2-4x-6y+1=0 \] is

• A. \((2,1)\)
• B. \((1,1)\)
• C. \((3,-1)\)
• D. \((3,1)\)

Hint:

For a family of polars involving a parameter \(\lambda\), collect all \(\lambda\)-terms together. If a point lies on every member of the family, then both the coefficient of \(\lambda\) and the constant part must be zero.

Answer 1

The Correct Option is D

Concept: For the circle \[ x^2+y^2+2gx+2fy+c=0, \] the polar of the point \[ (x_1,y_1) \] is \[ xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0. \] A point that lies on the polar for every value of a parameter must satisfy the resulting equation identically in that parameter.

Step 1: Identify the circle parameters. Given circle \[ x^2+y^2-4x-6y+1=0. \] Comparing with \[ x^2+y^2+2gx+2fy+c=0, \] we get \[ g=-2, \qquad f=-3, \qquad c=1. \]

Step 2: Write the equation of the polar of \((2\lambda,\lambda-4)\). Using \[ x_1=2\lambda, \qquad y_1=\lambda-4, \] the polar is \[ 2\lambda x+(\lambda-4)y -2(x+2\lambda) -3(y+\lambda-4) +1=0. \] Expanding, \[ 2\lambda x+\lambda y-4y -2x-4\lambda -3y-3\lambda+12+1=0. \] \[ \lambda(2x+y-7) -2x-7y+13=0. \]

Step 3: Find the point common to all such polars. Since the point lies on the polar for every real value of \(\lambda\), \[ \lambda(2x+y-7) -2x-7y+13 \equiv 0. \] Therefore, \[ 2x+y-7=0, \] and \[ -2x-7y+13=0. \]

Step 4: Solve the two equations. From \[ 2x+y=7, \] \[ y=7-2x. \] Substituting into \[ 2x+7y=13, \] \[ 2x+7(7-2x)=13. \] \[ 2x+49-14x=13. \] \[ -12x=-36. \] \[ x=3. \] Hence, \[ y=7-6=1. \]

Step 5: Write the final answer. \[ \boxed{(3,1)} \]