Question

For the given circles $x^2+y^2-6x-2y+1=0$ and $x^2+y^2+2x-8y+13=0$, which of the following is true?

• A. One circle lies inside the other
• B. One circle lies completely outside the other
• C. Two circle intersect in two points
• D. They touch each other externally

Answer 1

The correct option is D They touch each other externally

Given circles can be rewritten as

$(x-3{)}^2+(y-1{)}^2=9$ and $(x+1{)}^2+(y-4{)}^2=4$,

The centres of given circles are $C_1(3,1)$ and $C_2(-1,4)$ and corresponding radii are

$r_1=\sqrt{(3{)}^2+(1{)}^2-1}=3$

and $r_2=\sqrt{(-1{)}^2+(4{)}^2-13}=2$

Now, $C_1{C}_2=\sqrt{(-1-3{)}^2+(4-1{)}^2}=5$

$\therefore C_1{C}_2=r_1+r_2$

Hence, two circles touch externally.