The center of a cricle ids (h,7) and the radius is 10. The circle passes through (3,-1). Find all possible values of h.

2 answers:

8 0

$equation\ of\ a\ circle:\\\\(x-h)^2+(y-7)^2=10^2\\\\if\ \ (x;y)=(3;-1),\ then\ \ (3-h)^2+(-1-7)^2=100\\\\(3-h)^2+64=100\\\\ (3-h)^2=36\ \ \ \Leftrightarrow\ \ \ (3-h=6\ \ \ or\ \ \ 3-h=-6)\\\\3-h=6\ \ \ \ \ \Rightarrow\ \ \ h=-3\\3-h=-6\ \ \ \Rightarrow\ \ \ h=9\\\\Ans.\ h=-3\ \ or\ \ h=9$

Nostrana [21]3 years ago

6 0

$center\ of\ a\ circle:(a;\ b)\\\\radius:r\\\\(x-a)^2+(y-b)^2=r^2\\\\========================\\\\(h;\ 7);\ r=10\\\\(x-h)^2+(y-7)^2=10^2\\\\========================$

$The\ circle\ passes\ throught\ (3;-1).\\\\Substitute\ x=3\ and\ y=-1:\\\\(3-h)^2+(-1-7)^2=100\\\\(3-h)^2+(-8)^2=100\\\\(3-h)^2+64=100\ \ \ /-64\\\\(3-h)^2=36\iff3-h=\pm\sqrt{36}\\\\3-h=-6\ or\ 3-h=6\\-h=-6-3\ or\ -h=6-3\\-h=-9\ or\ -h=3\\h=9\ or\ h=-3\\\\Answer:h=9\ or\ h=-3$

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Which shows the following expression after the negative exponents have been eliminated? m^7 n^3/ mn^-1 assume m=0 and n=0

aliina [53]

Given expression is $\frac{m^7n^3}{mn^{-1}}$ .

where $m \neq 0, n \neq 0,$

Now we have to simplify this and find the next step so that negative exponent gets eliminated.

We know that when negative exponent changes sign, then it moves from numerator to denominator or from denominator to numerator as shown in the formula:

$x^{-m}=\frac{1}{x^m}$