Question

Show all work for the following

Answer 1

Answer:

$(-2-2\sqrt{2},3+2\sqrt{2})\,,\,(-2+2\sqrt{2},3-2\sqrt{2})$

Step-by-step explanation:

Given:

$(x+2)^2+(y-3)^2=16\\x+y-1=0$

To find: value of $x,y$

Solution:

$(x+2)^2+(y-3)^2=16\,\,\,...(i)\\x+y-1=0\,\,\,(ii)$

From (ii),

$x=1-y$

Put this value of $x$ in (i)

$(1-y+2)^2+(y-3)^2=16\\(3-y)^2+(y-3)^2=16\\(y-3)^2+(y-3)^2=16\\2(y-3)^2=16\\(y-3)^2=8$

$y-3=$ ± $\sqrt{8}$ = ± $2\sqrt{2}$

$y=3$ ± $2\sqrt{2}$

At $y=3$ + $2\sqrt{2}$ ,

$x=1-(3+2\sqrt{2})=1-3-2\sqrt{2}=-2-2\sqrt{2}$

At $y=3-2\sqrt{2}$ ,

$x=1-(3-2\sqrt{2})=1-3+2\sqrt{2}=-2+2\sqrt{2}$

Solutions are $(-2-2\sqrt{2},3+2\sqrt{2})\,,\,(-2+2\sqrt{2},3-2\sqrt{2})$