Question

The locus of the mid-points of the chords of the circle \( x^{2}+y^{2}-2 x-4 y-11=0 \) which subtend \( 60^{\circ} \) at the centre is 

(1) \( x^{2}+y^{2}-4 x-2 y-7=0 \) 

(2) \( x^{2}+y^{2}+4 x+2 y-7=0 \) 

(3) \( x^{2}+y^{2}-2 x-4 y-7=0 \) 

(4) \( x^{2}+y^{2}+2 x+4 y+7=0 \)

Comments

Correct answer is:- (3) x² + y² - 2x - 4y - 7 = 0

Explanation:-

Let AB be the chord of the circle and P be the midpoint of AB.

It is known that perpendicular from the center bisects a chord.

Thus, △ACP is a right-angled triangle. 

Now AC=BC= r (Radii) 

The equation of the given circle can be written as

(X - 1)² + (Y - 2)² = 16 

Hence, centre C=(1,2) and radius, r = 4 units. 

PC = AC sin60° 

= r sin60° 

= 4× √3/2 = 2√3 units

Therefore, PC = 2√3 units 

⇒ PC² = 12 

⇒ (X - 1)² + (Y - 2)² = 12

⇒ X² + Y²  - 2X - 4Y + 5 = 12

∴ X²  + Y² - 2X - 4Y - 7 = 0

Answer 1

The correct option is (3) x² + y² − 2x − 4y − 7 = 0.

Let AB be the chord of the circle and P be the midpoint of AB.

It is known that perpendicular from the center bisects a chord.

Thus △ACP is a right-angled triangle.

Now AC = BC = radius.

The equation of the give circle can be written as (x−1)² + (y−2)² = 16

Hence, centre C =(1, 2) and radius r = 4 units.