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Vinvika [58]
3 years ago
14

Write the equation of a circle with a center at (-1,2) and a radius of 7

Mathematics
1 answer:
Sever21 [200]3 years ago
5 0

Answer:

(x+1)^2+(y-2)^2=49

Step-by-step explanation:

The standard form for the equation of a circle is (x-h)^2+(y-k)^2=r^2 where  r  is the radius, and the center is (h, k).

(h, k) = (-1, 2),  r = 7

(x-(-1))^2+(y-2)^2=7^2\\\\(x+1)^2+(y-2)^2=49

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Given: cos θ=-4/5, sin x = -12/13, θ is in the third quadrant, 
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By definition of tangent,

tan(2<em>θ</em>) = sin(2<em>θ</em>) / cos(2<em>θ</em>)

Recall the double angle identities:

sin(2<em>θ</em>) = 2 sin(<em>θ</em>) cos(<em>θ</em>)

cos(2<em>θ</em>) = cos²(<em>θ</em>) - sin²(<em>θ</em>) = 2 cos²(<em>θ</em>) - 1

where the latter equality follows from the Pythagorean identity, cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1. From this identity we can solve for the unknown value of sin(<em>θ</em>):

sin(<em>θ</em>) = ± √(1 - cos²(<em>θ</em>))

and the sign of sin(<em>θ</em>) is determined by the quadrant in which the angle terminates.

<em />

We're given that <em>θ</em> belongs to the third quadrant, for which both sin(<em>θ</em>) and cos(<em>θ</em>) are negative. So if cos(<em>θ</em>) = -4/5, we get

sin(<em>θ</em>) = - √(1 - (-4/5)²) = -3/5

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tan(2<em>θ</em>) = sin(2<em>θ</em>) / cos(2<em>θ</em>)

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tan(2<em>θ</em>) = 24/7

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