Question

Find the position of the circles $x^2+y^2-2x-6y+9=0and{x}^2+y^2+6x-2y+1=0$ with respect to each other.

• A. Touch each other internally
• B. Touch each other externally
• C. One circle lies completely outside the other circle
• D. One circle lies completely inside the other circle
• E. Intersect each other at two points

Answer 1

The correct option is C

One circle lies completely outside the other circle

Given circles,

$x^2+y^2-2x-6y-9=0$ -------------(1)

$x^2+y^2+6x-2y+1=0$ --------------(2)

Let $c_1\&c_2$ be the centers and $r_1\&r_2$ radii of the circles (1) and (2) respectively.

Let $c_1(1,3)$

$c_2(-3,1)$

$c_1.c_2=\sqrt{(1+3{)}^2+(3-1{)}^2}=\sqrt{16+4}=2\sqrt{5}=2x2.23$

$=4.4721$

$r_1=\sqrt{g^2+f^2-c}=\sqrt{1+9-9}=1$

$r_2=\sqrt{g^2+f^2-c}=\sqrt{9+1-1}=3$

and $r_1+r_2=1+3=4$

so, we observe that $c_1{c}_2>r_1+r_2$

Hence. one circles lies completely outside the other circle.

Option C is correct.