Find The Locus Of The Mid Point Of The Chord Of The Circle x2 + y2 = a2which Subtend a right Angle At The Point (p, q).

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Question

Find the locus of the mid point of the chord of the circle x2 + y2 = a2which subtend a right angle at the point (p, q).

Solution

Correct option is

 

Let the mid point of the chord be M (h, k) and circle be represented by 

                   

Then equation of chord is T = S1

                hx + ky – a2 = h2 + k2 – a2

or            hx + ky = h2 + k2                       …(1)

 

OM = length of perpendicular from O (0, 0) to the line (1)

                                  

    

Since M is the mid point of AB

∴ MA = MB = MP (since distances from the mid point of hypotenuse of right angled triangle of the vertices are equal)

   

⇒             a2 – h2 – k2 = (h – p)2 + (k – q)2  

⇒              2h2 + 2k2 – 2ph – 2qk + p2 + q2 – a2 = 0

Hence locus of mid point M (h, k) is

                 2x2 + 2y2 – 2px – 2qy + p2 + q2 – a2 = 0  

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