If the circles x2+y2+2gx+2fy=0, and x2+y2+2g1x+2f1y=0 touch each other, ... r are their corresponding radii then if the circles touch each other, 1 answer
Answers
Answer: Two circles touch other if the line joining their centers is perpendicular to all ... circles x2+y2+2gx+2fy=0 and x2+y2+2g′x+2f′y=0 touch each other, then ...
Step-by-step explanation:
Answer:
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