Question

Find the point(s) of intersection of the circle with equation $x^2+y^2=4$ and the circle with equations ${(x-2)}^2+{(y-2)}^2=4$

• A. $(-2,0)$ and $(0,-2)$
• B. $(2,0)$ and $(0,2)$
• C. $(3,0)$ and $(0,3)$
• D. $(1,0)$ and $(0,1)$

Answer 1

The correct option is B $(2,0)$ and $(0,2)$

Let $x^2+y^2=4$ ...........(1)

We first expand the given second equation $(x-2{)}^2+(y-2{)}^2=4$ as follows:

$(x-2{)}^2+(y-2{)}^2=4$

$\Rightarrow x^2+4-4x+y^2+4-4y=4$

$\Rightarrow x^2+y^2-4x-4y=-4$ ........(2)

Now subtracting equation (1) from equation (2) we get,

$x^2+y^2-4x-4y-x^2-y^2=-4-4$

$\Rightarrow -4x-4y=-8$

$\Rightarrow 4x+4y=8$

$\Rightarrow x+y=2$

$\Rightarrow x=2-y$

We now substitute $x$ by $2-y$ in the first equation to obtain

$(2-y{)}^2+y^2=4$

$\Rightarrow 4+y^2-4y+y^2=4$

$\Rightarrow 2y^2-4y=4-4$

$\Rightarrow 2y^2-4y=0$

$\Rightarrow 2y(y-2)=0$

$\Rightarrow 2y=0$ and $(y-2)=0$

$\Rightarrow y=0$ and $y=2$

Put $y=0$ in equation (1) that is :

$x^2+(0{)}^2=4$

$\Rightarrow x^2=4$

$\Rightarrow x=2$

Now put $y=2$ in equation (1) that is :

$x^2+(2{)}^2=4$

$\Rightarrow x^2+4=4$

$\Rightarrow x^2=4-4$

$\Rightarrow x^2=0$

$\Rightarrow x=0$

The two points of intersection of the two circles are given by,

$(2,0)$ and $(0,2)$