: Find the equations to all possible common tangents of the circles x2 + y2 - 2x - 6y + 6 = 0 and x²+y²= 1
Answers
Answer:
Step-by-step explanation:
he centres and radii of the circles are
C
1
(1,3) and r
1
=
1+9−9
=1.
C
2
(−3,1) and r
2
=
9+1−1
=3.
C
1
C
2
=
20
,r
1
+r
2
=4=
16
∴C
1
C
2
>r
1
+r
2
. Hence the circles are non-intersecting. Thus there will be four common tangents.
Transverse common tangents are tangents drawn from the point P which divides C
1
C
2
internally in the ratio of radii 1:3.
Co-ordinates of P are
(
1+3
1(−3)+3.1
,
1+3
1.1+3.3
) i.e. (0,
2
5
).
Direct common tangents are tangents drawn from the point Q which divides C
1
C
2
externally in the ratio 1:3.
(m
2
+1)
m.1−3+5/2
=1
or (m−
2
1
)
2
=m
2
+1
or −m−3/4=0 or 0m
2
−m−3/4=0.
Hence m=−3/4 and ∞ as coeff. of m
2
is zero.
Therefore from (1),
x
y−5/2
=m=∞ and −3/4.
∴x=0 is a tangent and y−5/2=−3x/4
or 3x+4y−10=0
drawn from the point Q(3,4).
Now proceeding as for transverse tangents their equations are
y=4,4x−3y=0.