Circle and semi circle question (1 Viewer)

kpad5991 · Jan 18, 2017

I know how to sketch any circle however the question says sketch the coordinates of any points that have integer coordinates. How would you work that out for the following

x^2 + y^2 = 5

And y = root 2-x^2

eyeseeyou · Jan 18, 2017

kpad5991 said:
I know how to sketch any circle however the question says sketch the coordinates of any points that have integer coordinates. How would you work that out for the following

x^2 + y^2 = 5

And y = root 2-x^2

x^2+y^2=5 is common sense

Use the rule that x^2+y^2=r^2

r^2=5

therefore r^2= root 5

Draw a circle with the origin being the centre and draw a circle labelling all the necessary coordinates

https://www.google.com.au/search?q=...hVF2LwKHRu9D_oQ_AUIBigB#imgrc=arl4uR3WItosDM:

As for the second question I need you to be clearer, is it y= root (2-x^2) or y=root (2) -x^2 ?

pikachu975 · Jan 18, 2017

eyeseeyou said:
x^2+y^2=5 is common sense

Use the rule that x^2+y^2=r^2

r^2=5

therefore r^2= root 5

Draw a circle with the origin being the centre and draw a circle labelling all the necessary coordinates

https://www.google.com.au/search?q=...hVF2LwKHRu9D_oQ_AUIBigB#imgrc=arl4uR3WItosDM:

As for the second question I need you to be clearer, is it y= root (2-x^2) or y=root (2) -x^2 ?

You didn't even answer the question, he asked how do you find the INTEGER coordinates and all you did was tell him that the radius was irrational (root 5) which didn't help at all.

eyeseeyou · Jan 18, 2017

pikachu975 said:
You didn't even answer the question, he asked how do you find the INTEGER coordinates and all you did was tell him that the radius was irrational (root 5) which didn't help at all.

Whoops yeah soz idk

1729 · Jan 18, 2017

kpad5991 said:
I know how to sketch any circle however the question says sketch the coordinates of any points that have integer coordinates. How would you work that out for the following

x^2 + y^2 = 5

And y = root 2-x^2

$$\noindent For $x^{2} + y^{2} = 5$ it becomes obvious that the only two numbers $n_{1}, n_{2}$ such that $n_{1}, n_{2}\in\mathbb{Z}$ and $(n_{1})^{2} + (n_{2})^{2} = 5$ are $n_{1} = \pm1, \pm2$ and $n_{2} = \pm2, \pm1$ respectively \\\\ From this, we obtain the set of coordinates \\$\left \{(1,2);(2,1);(-1,2);(-2,1);(-1,-2);(-2,-1);(1,-2);(2,-1)\right \}$, where two coordinates exist in each quadrant.$\\\\$For $y = \sqrt{2-x^{2}}$ the domain, assuming $y = f(x)$, is $D_{f} = \left \{x \in \mathbb{R} \mid -\sqrt{2} \leqslant x \leqslant \sqrt{2}\right \}$ Within this domain, the only abscissae that gives an integer value of $y$ is $x = \pm 1$, thus we have the coordinates $\left \{(-1, 1);(1,1)\right \}$

pikachu975 · Jan 18, 2017

eyeseeyou said:
x^2+y^2=5 is common sense

Use the rule that x^2+y^2=r^2

r^2=5

therefore r^2= root 5

Draw a circle with the origin being the centre and draw a circle labelling all the necessary coordinates

https://www.google.com.au/search?q=...hVF2LwKHRu9D_oQ_AUIBigB#imgrc=arl4uR3WItosDM:

As for the second question I need you to be clearer, is it y= root (2-x^2) or y=root (2) -x^2 ?

Just a note, the whole equation is under square root as a semicircle equation is sqrt(r^2 - x^2)

Circle and semi circle question (1 Viewer)

kpad5991

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eyeseeyou

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pikachu975

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eyeseeyou

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1729

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pikachu975

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