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Sati [7]
3 years ago
15

What is the correct standard equation of a circle with center (5,-1) and radius 12

Mathematics
1 answer:
Charra [1.4K]3 years ago
6 0

The equation for a circle is

(x-h)² + (y-k)² = r²

h = your given x

y = your given k

Let's plug everything in!

(x - 5) ² + (y - (-1)) ² = 12²

Your final equation is

(x - 5) ² + (y + 1) ² = 144

<em>Hope I helped! Comment or message me if you have any questions :) </em>

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\purple{ \bold{ \tan( \alpha  -  \beta ) = 1.00701798}}

Step-by-step explanation:

\cos( \alpha ) =  \frac{ \sqrt{6} }{6}  =  \frac{1}{ \sqrt{6} }  \\  \\  \therefore \:  \sin( \alpha )  =  \sqrt{1 -  { \cos}^{2} ( \alpha ) }  \\  \\  =  \sqrt{1 -  \bigg( {\frac{1}{ \sqrt{6} } \bigg )}^{2} }  \\  \\ =  \sqrt{1 -  {\frac{1}{ {6} }}}  \\  \\ =  \sqrt{ {\frac{6 - 1}{ {6} }}}   \\  \\  \red{\sin( \alpha ) =  \sqrt{ { \frac{5}{ {6} }}} } \\  \\  \tan( \alpha ) =  \frac{\sin( \alpha ) }{\cos( \alpha ) }  =  \sqrt{5}  \\  \\ \sin( \beta )  =  \frac{ \sqrt{2} }{4}  \\  \\  \implies \: \cos( \beta )  =   \sqrt{ \frac{7}{8} }  \\  \\ \tan( \beta )  =  \frac{\sin( \beta ) }{\cos( \beta ) } =  \frac{1}{ \sqrt{7} }   \\  \\  \tan( \alpha  -  \beta ) =  \frac{ \tan \alpha  -  \tan \beta }{1 +  \tan \alpha .  \tan \beta}  \\  \\  =  \frac{ \sqrt{5} -  \frac{1}{ \sqrt{7} }  }{1 +  \sqrt{5} . \frac{1}{ \sqrt{7} } }  \\  \\  =  \frac{ \sqrt{35} - 1 }{ \sqrt{7}  +  \sqrt{5} }  \\  \\  \purple{ \bold{ \tan( \alpha  -  \beta ) = 1.00701798}}

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Step-by-step explanation:

Hello! Thanks for letting me answer your question! If you have any more questions feel free to ask me! If not, Have a WONDERFUL day and Please don't forget to consider making me Brainliest on your question today!

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