If the radical axis of the circles x2 + y2 + 2gx + 2fy + c = 0 and 2x2 + 2y2 + 3x + 8y + 2c = 0 touches the circle x2 + y2 + 2x + 2y + 1 = 0 , then (4g - 3)(f - 2) =

Question

If the radical axis of the circles x2 + y2 + 2gx + 2fy + c = 0 and 2x2 + 2y2 + 3x + 8y + 2c = 0 touches the circle x2 + y2 + 2x + 2y + 1 = 0 , then (4g - 3)(f - 2) = (A) 0 (B) -1 (C) 1 (D) 2

Tardigrade Admin · Accepted Answer

The radical axis of the circles x2+y2+2 g x+2 f y+ c=0 and 2 x2+2 y2+3 x+8 y+2 c=0 is (4 g-3) x+(4 f-8) y=0 ...(i) Since, the radical axis (i) touches the circle x2+y2+ 2 x+2 y+1=0, so (|-(4 g-3)-(4 f-8)|/√(4 g-3)2+(4 f-8)2)=√1+1-1 ⇒ (4 g-3)2+(4 f-8)2+2(4 g-3)(4 f-8) =(4 g-3)2+(4 f-8)2 ⇒ 8(4 g-3)(f-2)=0 ⇒ (4 g-3)(f-2)=0

Q. If the radical axis of the circles x2+y2+2gx+2fy+c=0 and 2x2+2y2+3x+8y+2c=0 touches the circle x2+y2+2x+2y+1=0 , then (4g−3)(f−2)=

0

-1

1

2

Q. If the radical axis of the circles $x^{2} + y^{2} + 2 gx + 2 f y + c = 0$ and $2 x^{2} + 2 y^{2} + 3 x + 8 y + 2 c = 0$ touches the circle $x^{2} + y^{2} + 2 x + 2 y + 1 = 0$ , then $(4 g - 3) (f - 2) =$