1
Question
If the circle x2+y2+2gx+2fy+c=0 cuts the three circles x2+y2−5=0,x2+y2−8x−6y+10=0 and x2+y2−4x+2y−2=0 at the extremities of their diameters, then
If the circle x2+y2+2gx+2fy+c=0 cuts the three circles x2+y2−5=0,x2+y2−8x−6y+10=0 and x2+y2−4x+2y−2=0 at the extremities of their diameters, then
Open in App
Solution
The correct options are
A fg=147/25
B 4f=3g
D C=−5
Since x2+y2+2gx+2fy+c=0 cuts the 3 circles at the ends of their diameters, the
common chords are diameters.
⟹
Centers pass through their respective common chords.
Common chords
are given by
S1–S2=0
For first circle,
2gx+2fy+c+5=0(1)
For second circle,
(2g+8)x+(2f+6)y+(c–10)=0(2)
For third circle,
(2g+4)x+(2f−2)y+(c+2)=0(3)
C1(0,0)
passes through (1)
⟹c+5=0⟹c=−5
C2(4,3)
passes through (2)
⟹8g+6f=−35–(4)
C3(−2,1)
passes through (3)
⟹4g–2f=−7–(5)
Solving (4) and
(5)
f=−2110,g=−145
fg=14725
g+2f=−7
c+2=−5+2=−3
g+2f≠c+2
4f=−425=3g
A fg=147/25
B 4f=3g
D C=−5
Since x2+y2+2gx+2fy+c=0 cuts the 3 circles at the ends of their diameters, the common chords are diameters.
⟹ Centers pass through their respective common chords.
Common chords are given by
S1–S2=0
For first circle,
2gx+2fy+c+5=0(1)
For second circle,
(2g+8)x+(2f+6)y+(c–10)=0(2)
For third circle,
(2g+4)x+(2f−2)y+(c+2)=0(3)
C1(0,0) passes through (1)
⟹c+5=0⟹c=−5
C2(4,3) passes through (2)
⟹8g+6f=−35–(4)
C3(−2,1) passes through (3)
⟹4g–2f=−7–(5)
Solving (4) and (5)
f=−2110,g=−145
fg=14725
g+2f=−7
c+2=−5+2=−3
g+2f≠c+2
4f=−425=3g
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
