1
:x
2
+y
2
−6x−9y+13=0
(x−3)
2
+(y−
2
9
)
2
−9−
4
81
+13=0
(x−3)
2
+(y−
2
9
)
2
=
4
65
Here,
r
1
=
2
65
C
1
=(3,
2
9
)
Equation of another circle-
S
2
:x
2
+y
2
−2x−16y=0
(x−1)
2
+(y−8)
2
−1−64=0
(x−1)
2
+(y−8)
2
=65
Here,
r
2
=
65
C
2
=(1,8)
Distance between the centre of two circles-
C
1
C
2
=
(3−1)
2
+(8−
2
9
)
2
C
1
C
2
=
4+
4
49
=
2
65
∣r
2
−r
1
∣=
∣
∣
∣
∣
∣
∣
65
−
2
65
∣
∣
∣
∣
∣
∣
=
2
65
∵C
1
C
2
=∣r
1
−r
2
∣
Thus the two circles touches each other internally.
Since the circle touches each other internally. The point of contact P divides C
1
C
2
externally in the ratio r
1
:r
2
, i.e.,
2
65
:
65
=1:2
Therefore, coordinates of P are-
⎝
⎜
⎜
⎜
⎜
⎜
⎛
1−2
1(1)−2(3)
,
1−2
1(8)−2(
2
9
)
⎠
⎟
⎟
⎟
⎟
⎟
⎞
=(5,1)
Therefore,
Equation of common tangent is-
S
1
−S
2
=0
(5x+y−6(
2
x+5
)−9(
2
y+1
)+13)−(5x+y−2(
2
x+5
)−16(
2
y+1
))=0
2
−6x−9y−13
+x+8y+13=0
4x−7y−13=0
Hence the point of contact is (5,1) and the equation of common tangent is 4x−7y−13=0.
Answer:
6:5
Step-by-step explanation:
Bubble algae cells can grow up to 5 cm in diameter. Calculate the surface area : volume ratio of this cell.
The algae is spherical in nature
Volume of the algae =4/3 pi r³
Volume of the algae = 4/3×pi×2.5³
Volume of the algae = 62.5/3 pi
Surface area = 4pi r²
Surface area = 4pi ×2.5²
Surface area = 25pi
Taking their ratio
surface area : volume ratio
= 25pi:62.5/3 pi
= 25 × 3/62.5
= 1.2
Hence the required ratio is 12/10 which is 6:5
Answer:
-5c
Step-by-step explanation:
c+c-7c > 2c-7c = -5c
Answer:
Note that
0.16666...is equivalent to 1/6
Step-by-step explanation:
We can use the sum of fractions to write out the decimal:
For this particular question, we can see that
0.16666...is the fraction 1/6, therefore
the given decimal
2.16666...=2+1/6=13/6
Answer:
Step-by-step explanation:
time to cross will be the distance divided by the velocity in that direction.
a) t = d/v = 45/2.75 = 16.363636... 16.4 s
In the same time that the otter is swimming cross stream, the stream is also moving perpendicularly at its own pace.
b) d = vt = 2.5(16.36) = 40.9090... 40.9 m
As velocity is a vector, An observer on the bank would see the velocity as the vector addition of the cross stream and down stream velocities
c) v = √(2.75² + 2.5²) = 3.716517... 3.72 m/s magnitude
θ = arctan2.5/2.75 = 42.2736... 42.3° downstream of a line straight across
