Should be 0 if this is exponents
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.
9514 1404 393
Answer:
9.5°, yes
Step-by-step explanation:
The relevant trig relation is ...
Tan = Opposite/Adjacent
The distance opposite the angle of elevation is the plane's height, 500 m. The distance adjacent to the angle of elevation is the horizontal distance to the plane, 3 km = 3000 m. Then the angle is found from ...
tan(α) = 500/3000 = 1/6
α = arctan(1/6) ≈ 9.46°
The plane is approaching at an angle of 9.46°. It is safe to land, since that angle is less than 15°.
_____
<em>Additional comment</em>
The usual descent angle for most commercial air traffic is 3°. Some airport geography demands it be different (steeper). A higher descent angle can put undue stress on the landing gear.
What does r mean. if r means multiply blank than 3*4 = 12
