Question

The circles $x^2+y^2+2x-2y+1=0and{x}^2+y^2-2x-2y+1=0$

• A. touch each other externally
• B. touch each other internally
• C. intersect on the y-axis
• D. touch each other at the point $(0,1).$

Answer 1

Answer: (A), (D)

touch each other at the point $(0,1).$
touch each other externally

$S_2-S_1=0$ implies $(x^2+y^2+2x-2y+1)-(x^2+y^2-2x-2y+1)=0$

$\therefore 4x=0$

$\therefore x=0$

Substituting in $x^2+y^2-2x-2y+1=0$, we get

$(y-1{)}^2=0$
$y=1$.

Now the center to center distance of the circles is

$=\sqrt{(1-(-1){)}^2+(1-1{)}^2}$
$=2$.

And $r_1+r_2$
$=1+1$
$=2$

Therefore,$r_1+r_2=C_1{C}_2$.

Hence the circles touch each other, and since the centers are $(-1,1)$ and $(1,1)$, hence they touch each other externally at $(0,1)$.