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2 years ago

11

What is the radius of a circle whose equation is x2 + (y – 8)2 = 25?

1 answer:

Nesterboy [21]2 years ago

6 0

The formula for a circle is (x - h)^2 + (y - k)^2 = r^2
Where (h, k) is the center and r is the radius.
x^2 + (y - 8)^2 = 25
25 = r^2
So the radius is sqrt 25 or 5.
Radius = 5

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$2\sin\alpha-\cos\alpha=2$

Consider the substitution $\tan\dfrac\alpha2=\beta$ . Then by the double angle identities we get

$\sin\alpha=2\sin\dfrac\alpha2\cos\dfrac\alpha2$

$\cos\alpha=\cos^2\dfrac\alpha2-\sin^2\dfrac\alpha2$

We also have

$\tan\dfrac\alpha2=\beta\implies\begin{cases}\sin\dfrac\alpha2=\dfrac\beta{\sqrt{1+\beta^2}}\\\\\cos\dfrac\alpha2=\dfrac1{\sqrt{1+\beta^2}}\end{cases}$

so that

$\sin\alpha=\dfrac{2\beta^2}{1+\beta^2}$

$\cos\alpha=\dfrac{1-\beta^2}{1+\beta^2}$

and the original equation has been transformed to

$\dfrac{4\beta^2-(1-\beta^2)}{1+\beta^2}=2$

Solve for $\beta$ :

$5\beta^2-1=2+2\beta^2$

$3\beta^2=3$

$\beta^2=1$

$\beta=\pm1$

Solving for $\alpha$ gives

$\tan\dfrac\alpha2=-1\implies\dfrac\alpha2=-\dfrac\pi4+n\pi\implies\alpha=-\dfrac\pi2+2n\pi$

$\tan\dfrac\alpha2=1\implies\dfrac\alpha2=\dfrac\pi4+n\pi\implies\alpha=\dfrac\pi2+2n\pi$

where $n$ is any integer. Both $\sin$ and $\cos$ are $2\pi$ -periodic, which is to say

$\cos(x+2n\pi)=\cos x$

$\sin(x+2n\pi)=\sin x$

so that

$\sin\alpha=\sin\left(\pm\dfrac\pi2+2n\pi\right)=\sin\left(\pm\dfrac\pi2\right)=\pm1$

$\cos\alpha=\cos\left(\pm\dfrac\pi2+2n\pi\right)=\cos\left(\pm\dfrac\pi2\right)=0$

and we find that

$\sin\alpha+2\cos\alpha=\pm1$

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