Question

If A = {(x, y)/x² + y² =54; x,y belong toR} and B={(x,y)/x2 + y2 29; x, y belong to R}, then
1) A-B= phi
2) B-A=phi
3) A intersection B is not equal to phi
4) A intersection B = phi​

Answer 1

Answer:

We have R={(x,y):x,y∈Z,x

2

+y

2

≤4}.

Let x=0∴x

2

+y

2

≤4⇒y

2

≤4⇒y=0,±1,±2

Let x=±1∴x

2

+y

2

≤4⇒y

2

≤3⇒y=0,±1

Let x=±2∴x

2

+y

2

≤4⇒y

2

≤0⇒y=0

∴R={(0,0),(0,−1),(0,1),(0,−2),(0,2),(−1,0),(1,0),(1,1),(1,−1),(−1,1),(2,0),(−2,0)}

∴ Domain of R={x:(x,y)ϵR}={0,−1,1,−2,2}.

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Answer 2

$\large\text{\underline{Note}}$

Please check your question once.

I am assuming set B as $B=\{(x,y)|x^{2}+y^{2}>29\}$.

$\large\text{\underline{Solution}}$

A is a circle of a radius of $\sqrt{54}$, centered at the origin, and B is the outside area of the circle of a radius of $\sqrt{29}$, centered at the origin.

As $A\subset B$, the correct answer will be $A-B=\varnothing$ and $A\cap B\neq\varnothing$.

$\large\text{\underline{Conclusion}}$

So, the correct options are 1) and 3).