Question

if two circles s=xsquare+ysquare+2gx+2fy+c=0 , spower1= xsquare+ysquare+2gpower1x+2fpower1y+cpower1=0 cut each other orthoganally then condition for orthoganalty is​

Answer 1

Answer:

$\huge{\textbf{\textsf{{\color{navy}{Ans}}{\purple{wer}}{\pink{♧}}{\color{pink}{᭄}}}}}$

f1g=fg1

If C1 and C2 are the centres of the circles respectively and R and r are their corresponding radii then if the circles touch each other,

C1C2=R+r

Consider equation 1

(x+g)2+(y+f)2=g2+f2

Similarly,

(x+g1)2+(y+f1)2=(g1)2+(f1)2

Therefore 

C1C2=R+r implies,

(g−g1)2+(f−f1)2=g2+f2+(g1)2+(f1)2

Squaring both sides

(g−g1)

Answer 2

Answer

Correct option is

B

f

1

g=fg

1

If C

1

and C

2

are the centres of the circles respectively and R and r are their corresponding radii then if the circles touch each other,

C

1

C

2

=R+r

Consider equation 1

(x+g)

2

+(y+f)

2

=g

2

+f

2

Similarly,

(x+g

1

)

2

+(y+f

1

)

2

=(g

1

)

2

+(f

1

)

2

Therefore

C

1

C

2

=R+r implies,

(g−g

1

)

2

+(f−f

1

)

2

=

g

2

+f

2

+

(g

1

)

2

+(f

1

)

2

Squaring both sides

(g−g

1

)

2

+(f−f

1

)

2

=g

2

+(g

1

)

2

+f

2

+(f

1

)

2

+2

g

2

+f

2

.

(g

1

)

2

+(f

1

)

2

g

2

+(g

1

)

2

+f

2

+(f

1

)

2

−2gg

1

−2ff

1

=g

2

+(g

1

)

2

+f

2

+(f

1

)

2

+2

g

2

+f

2

.

(g

1

)

2

+(f

1

)

2

−2gg

1

−2ff

1

=2

g

2

+f

2

.

(g

1

)

2

+(f

1

)

2

−gg

1

−ff

1

=

g

2

+f

2

.

(g

1

)

2

+(f

1

)

2

Squaring both sides, we get

(gg

1

)

2

+(ff

1

)

2

+2gg

1

ff

1

=(g

2

+f

2

).((g

1

)

2

+(f

1

)

2

)

(gg

1

)

2

+(ff

1

)

2

+2gg

1

ff

1

=(gg

1

)

2

+(ff

1

)

2

+(gf

1

)

2

+(fg

1

)

2

2gg

1

ff

1

=(gf

1

)

2

+(fg

1

)

2

(gf

1

)

2

+(fg

1

)

2

−2gg

1

ff

1

=0

(gf

1

−fg

1

)

2

=0

gf

1

=fg

1