Question

The locus of the centre of a circle touching the lines \[ x+2y=0 \] and \[ x-2y=0 \] is

• A. \(xy=0\)
• B. \(x=0\)
• C. \(y=0\)
• D. \(x+y=0\)

Hint:

The centre of a circle touching two intersecting lines always lies on one of their angle bisectors. Use \[ \boxed{\frac{L_1}{\sqrt{a_1^2+b_1^2}} = \pm \frac{L_2}{\sqrt{a_2^2+b_2^2}}} \] to find the angle bisectors.

Answer 1

The Correct Option is A

Step 1: Use the property of the centre. The centre of a circle touching two intersecting lines lies on the angle bisectors of the lines. The given lines are \[ x+2y=0 \] and \[ x-2y=0. \]

Step 2: Find the angle bisectors. The angle bisectors satisfy \[ \frac{x+2y}{\sqrt{1^2+2^2}} = \pm \frac{x-2y}{\sqrt{1^2+(-2)^2}}. \] Since both denominators are equal, \[ x+2y=\pm(x-2y). \] For the positive sign, \[ x+2y=x-2y \] gives \[ y=0. \] For the negative sign, \[ x+2y=-x+2y \] gives \[ x=0. \]

Step 3: Write the locus. Hence the locus is the pair of lines \[ x=0 \quad\text{or}\quad y=0, \] whose combined equation is \[ xy=0. \] Therefore, \[ \boxed{xy=0}. \] Thus, \[ \boxed{(A)} \] is the correct answer.