Find the equation of the circle passing through the origin and with the center (3,5)

1 answer:

4 0

$Equation\ of\ circle\\\\\
r^2=(x-a)^2+(y-b)^2\\\\
a,b- \ coordinates\ of\ center\ of\ circle\\\\\
(a,b)=(3,5)\\\\
r^2=(x-3)^2+(y-5)^2\\\\\
Substituting\ point\ (0,0)-origin\\\\
r^2=(0-3)^2+(0-5)^2\\\\
r^2=(-3)^2+(-5)^2\\\\
r^2=9+25=36\\
r=6\\\\
Answer:\ \ 36=(x-3)^2+(y-5)^2$

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What is the square root of -2i?

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Answer:

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Step-by-step explanation:

We are to find the square roots of $z=0-2i$ . First, convert from Cartesian to polar form:

$r=\sqrt{a^2+b^2}\\r=\sqrt{0^2+(-2)^2}\\r=\sqrt{0+4}\\r=\sqrt{4}\\r=2$

$\theta=tan^{-1}(\frac{b}{a})\\ \theta=tan^{-1}(\frac{-2}{0})\\\theta=\frac{3\pi}{2}$

$z=2(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2})$

Next, use the formula $\displaystyle \sqrt[n]{r}\biggr[\cis\biggr(\frac{\theta+2\pi k}{n}\biggr)\biggr]$ where $\displaystyle k=0,1,2,...\:,n-1$ to find the square roots: