Question

The locus of the mid points of the chords of the circle $x^2+y^2+4x-6y-12=0$ which subtend an angle of $\pi /3$ radians at its circumference is

• A. $(x+2{)}^2+(y-3{)}^2=6.25$
• B. $(x-2{)}^2+(y+3{)}^2=6.25$
• C. $(x+2{)}^2+(y-3{)}^2=18.75$
• D. $(x+2{)}^2+(y+3{)}^2=18.75$

Answer 1

The correct option is A $(x+2{)}^2+(y-3{)}^2=6.25$

Circle equation : $x^2+y^2-4x-6y-12=0$

angle $=\frac{\pi }{3}={60}^o$

centre $(-2,3)$

radius :

$p=5\cos \theta =5\cos \frac{\pi }{3}$

$=5\cos {60}^o=\frac{5}{2}$

$p=2.5$

locus $\Rightarrow (x-x_1{)}^2+(y-y_1{)}^2=6.25$

$(x+2{)}^2+(y-3{)}^2=6.25$

$\therefore (h,k)\Rightarrow (h+2{)}^2+(k-3{)}^2=6.25$

$(Option:A)$