Question

If the locus of midpoints of the chords of the circle \[ S\equiv x^2+y^2-6x-8y-11=0 \] which subtend a right angle at \(A(1,2)\) is another circle \(S'=0\), then the centre of \(S'=0\)

• A. is the midpoint of the line segment joining \(A\) and the centre of \(S=0\)
• B. Divides the line segment joining \(A\) and the centre of \(S=0\) in the ratio \(1:2\)
• C. Lies outside the circle \(S=0\)
• D. A vertex of the triangle having \(A\) and the centre of \(S=0\) as other two vertices

Hint:

For a fixed point \(P\), the locus of the midpoints of the chords of a circle subtending a right angle at \(P\) is another circle whose centre is the midpoint of the line joining \(P\) and the centre of the given circle.

Answer 1

The Correct Option is A

Step 1: Find the centre of the given circle. The given circle is \[ x^2+y^2-6x-8y-11=0. \] Comparing with the standard form, \[ (x-3)^2+(y-4)^2=36, \] its centre is \[ C(3,4). \]

Step 2: Use the standard result. The locus of the midpoints of chords of a circle which subtend a right angle at a fixed point is itself a circle. Its centre is the midpoint of the line segment joining the fixed point and the centre of the given circle. Here, \[ A=(1,2), \qquad C=(3,4). \] Hence the centre of the required circle is \[ \left(\frac{1+3}{2},\frac{2+4}{2}\right) =(2,3), \] which is the midpoint of \(AC\).

Step 3: Write the conclusion. Therefore, the centre of \[ S'=0 \] is the midpoint of the line segment joining \[ A \] and the centre of \[ S=0. \] Hence, \[ \boxed{\text{Option (A)}}. \] Thus, \[ \boxed{(A)} \] is the correct answer.