Question

The locus of the centre of the circles which touch both the circles x2+y2=A2 and x2+y2=4ax

Answer 1

C: x2+y2=a2P1: y2=4axP2: y2=−4ax So, x2+y2=a2 x2+(4ax)=a2 x2+4ax−a2=0using quadratic formulax=−b±b2−4ac√2a (where a=1,b=4a,c=−a2)x=−4a±16a2+4a2√2x=−4a±20a2√2x=−4a±2a5√2x=−2a±a5‾√For A , x>0 (on positive axis)hence x=−2a+a5 ‾‾√or x=a(5‾√−2)now put value of x in y2=4axy=4a2(5‾√−2)‾‾‾‾‾‾‾‾‾‾‾‾√Point A(a(5‾√−2),4a2(5‾√−2)‾‾‾‾‾‾‾‾‾‾‾‾√)Area of portion A is also rectangleso Area of A=a(5‾√−2))×4a2(5‾√−2)‾‾‾‾‾‾‾‾‾‾‾‾√area of whole rectangle is =4×portion of A                                                =4×a(5‾√−2)×4a2(5‾√−2)‾‾‾‾‾‾‾‾‾‾‾‾√                                                =4a(5‾√−2)×2a(5‾√−2)‾‾‾‾‾‾‾‾‾√                                                =8a2(5‾√−2)(5‾√−2)‾‾‾‾‾‾‾‾‾√                                                =8a2(5‾√−2)32Hence Option C is correct