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tia_tia [17]
3 years ago
15

Whats the equation of this Circle?

Mathematics
2 answers:
Leviafan [203]3 years ago
7 0

So the answer is this

-2+-7=-5

Kaylis [27]3 years ago
7 0

equation of a circle is given as

(x-a)^2 + (y-b)^2 = r^2

where (a,b) is coordinates of center of the circle and r is radius.

as per shown in graph we can see center of circle.is ( -7, -2)

now let's calculate radius. we have to calculate distance between any point on circle and center . let's take point (-7,0) on the circle shown in graph.

r = √((-7-(-7))^2 + (0-(-2))^2

r = √4 = 2

so equation is

(x -(-7))^2 + (y - (-2))^2 = 2^2

(x +7)^2 + (y +2)^2 = 4

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Prove that sinA-sin3A+sin5A-sin7A/cosA-cos3A-cos5A+cos7A= cot2A
MAVERICK [17]
Write the left side of the given expression as N/D, where
N = sinA - sin3A + sin5A - sin7A
D = cosA - cos3A - cos5A + cos7A
Therefore we want to show that N/D = cot2A.

We shall use these identities:
sin x - sin y = 2cos((x+y)/2)*sin((x-y)/2)
cos x - cos y = -2sin((x+y)/2)*sin((x-y)2)

N = -(sin7A - sinA) + sin5A - sin3A
    = -2cos4A*sin3A + 2cos4A*sinA
    = 2cos4A(sinA - sin3A)
    = 2cos4A*2cos(2A)sin(-A)
    = -4cos4A*cos2A*sinA

D = cos7A + cosA - (cos5A + cos3A)
   = 2cos4A*cos3A - 2cos4A*cosA
   = 2cos4A(cos3A - cosA)
   = 2cos4A*(-2)sin2A*sinA
   = -4cos4A*sin2A*sinA

Therefore
N/D = [-4cos4A*cos2A*sinA]/[-4cos4A*sin2A*sinA]
       = cos2A/sin2A
      = cot2A

This verifies the identity.
4 0
3 years ago
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