Answer:
[tex](-2-2\sqrt{2},3+2\sqrt{2})\,,\,(-2+2\sqrt{2},3-2\sqrt{2})[/tex]
Step-by-step explanation:
Given:
[tex](x+2)^2+(y-3)^2=16\\x+y-1=0[/tex]
To find: value of [tex]x,y[/tex]
Solution:
[tex](x+2)^2+(y-3)^2=16\,\,\,...(i)\\x+y-1=0\,\,\,(ii)[/tex]
From (ii),
[tex]x=1-y[/tex]
Put this value of [tex]x[/tex] in (i)
[tex](1-y+2)^2+(y-3)^2=16\\(3-y)^2+(y-3)^2=16\\(y-3)^2+(y-3)^2=16\\2(y-3)^2=16\\(y-3)^2=8[/tex]
[tex]y-3=[/tex] ± [tex]\sqrt{8}[/tex] = ± [tex]2\sqrt{2}[/tex]
[tex]y=3[/tex] ± [tex]2\sqrt{2}[/tex]
At [tex]y=3[/tex] + [tex]2\sqrt{2}[/tex] ,
[tex]x=1-(3+2\sqrt{2})=1-3-2\sqrt{2}=-2-2\sqrt{2}[/tex]
At [tex]y=3-2\sqrt{2}[/tex] ,
[tex]x=1-(3-2\sqrt{2})=1-3+2\sqrt{2}=-2+2\sqrt{2}[/tex]
Solutions are [tex](-2-2\sqrt{2},3+2\sqrt{2})\,,\,(-2+2\sqrt{2},3-2\sqrt{2})[/tex]
