(x+2)^2 + (y-5)^2 <16

Y >/= x^2 + 22
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1 Answer

1...(x+2)^2 +(y-5)^2 < 16 be a serkel with radius=4 & senter at (-2,5)

it sae...stuf inside the serkel

2... y>=x^2+22 be a parabola that open tu the rite...horizontal axis along y=22

Thats wae abuv the serkel...intersekt=null
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