Answer:
a is the value of X
Step-by-step explanation:
Answer:
![(x+8)^2 +(y-4)^2 = 25](https://tex.z-dn.net/?f=%20%28x%2B8%29%5E2%20%2B%28y-4%29%5E2%20%3D%2025)
Step-by-step explanation:
Assuming this complete problem: "A cell tower is located at (-8, 4) and transmits a circular signal that covers three major cities. The three cities are located on the circle and have the following coordinates: G (-4, 7), H (-13, 4), and I (-8, -1). Find the equation of the circle"
For this case the generla equation for the circle is given by:
![(x-h)^2 +(y-k)^2 = r^2](https://tex.z-dn.net/?f=%20%28x-h%29%5E2%20%2B%28y-k%29%5E2%20%3D%20r%5E2)
From the info we know that the tower is located at (-8, 4) so then h = -8 and k = 4, so then we need to find the radius. So we have the equation like this:![(x+8)^2 +(y-4)^2 = r^2](https://tex.z-dn.net/?f=%20%28x%2B8%29%5E2%20%2B%28y-4%29%5E2%20%3D%20r%5E2)
If the 3 points are on the circle then satisfy the equation. We can use the first point (-4,7) and if we replace we can find the value for ![r^2](https://tex.z-dn.net/?f=r%5E2)
![(-4+8)^2 +(7-4)^2 =25= r^2](https://tex.z-dn.net/?f=%20%28-4%2B8%29%5E2%20%2B%287-4%29%5E2%20%3D25%3D%20r%5E2)
So then ![r = \sqrt{25}=5](https://tex.z-dn.net/?f=%20r%20%3D%20%5Csqrt%7B25%7D%3D5)
And if we replace the second point we got this:
![(-13+8)^2 +(4-4)^2 = 25 =r^2](https://tex.z-dn.net/?f=%20%28-13%2B8%29%5E2%20%2B%284-4%29%5E2%20%3D%2025%20%3Dr%5E2)
And for the third point we have:
![(-8+8)^2 +(-1-4)^2 = 25 =r^2](https://tex.z-dn.net/?f=%20%28-8%2B8%29%5E2%20%2B%28-1-4%29%5E2%20%3D%2025%20%3Dr%5E2)
And we got the same result.
So then our final equation is given:
![(x+8)^2 +(y-4)^2 = 25](https://tex.z-dn.net/?f=%20%28x%2B8%29%5E2%20%2B%28y-4%29%5E2%20%3D%2025)
Check the picture below.
now, we're making an assumption that, the two blue shaded region are equal in shape, and thus if that's so, that area above the 14 is 6 and below it is also 6, 14 + 6 + 6 = 26.
so hmm if we simply get the area of the trapezoid and subtract the area of the yellow triangle and the area of the cyan triangle, what's leftover is what we didn't subtract, namely the shaded region.
![\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h~~=height\\ a,b=\stackrel{parallel~sides}{bases~\hfill }\\[-0.5em] \hrulefill\\ h=15\\ a=14\\ b=26 \end{cases}\implies A=\cfrac{15(14+26)}{2}\implies A=300 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Areas}}{\stackrel{trapezoid}{300}~~ - ~~\stackrel{yellow~triangle}{\cfrac{1}{2}(26)(9)}~~ - ~~\stackrel{cyan~triangle}{\cfrac{1}{2}(15)(6)}} \\\\\\ 300~~ - ~~117~~ - ~~45\implies 138\qquad \textit{blue shaded area}](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20trapezoid%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7Bh%28a%2Bb%29%7D%7B2%7D~~%20%5Cbegin%7Bcases%7D%20h~~%3Dheight%5C%5C%20a%2Cb%3D%5Cstackrel%7Bparallel~sides%7D%7Bbases~%5Chfill%20%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20h%3D15%5C%5C%20a%3D14%5C%5C%20b%3D26%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B15%2814%2B26%29%7D%7B2%7D%5Cimplies%20A%3D300%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Areas%7D%7D%7B%5Cstackrel%7Btrapezoid%7D%7B300%7D~~%20-%20~~%5Cstackrel%7Byellow~triangle%7D%7B%5Ccfrac%7B1%7D%7B2%7D%2826%29%289%29%7D~~%20-%20~~%5Cstackrel%7Bcyan~triangle%7D%7B%5Ccfrac%7B1%7D%7B2%7D%2815%29%286%29%7D%7D%20%5C%5C%5C%5C%5C%5C%20300~~%20-%20~~117~~%20-%20~~45%5Cimplies%20138%5Cqquad%20%5Ctextit%7Bblue%20shaded%20area%7D)