Question

Examine if the two circles$x^2+y^2-2x-4y=0and{x}^2+y^2-8y-4=0$ _________.

• 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 A

Touch each other internally

Given circles,

$x^2+y^2-2x-4y=0$ -------------(1)

$x^2+y^2-8x-4=0$ --------------(2)

Let $c_1\&c_2$ be the centers and $r_{1}and{r}_2$ the radius of circles (1) and (2) respectively. Then

Centers of the circles is (-g,f)

$c_1(1,2)$

$c_2(-0,4)$

radius $r_1=\sqrt{g^2+f^2-c}=\sqrt{1+4-0}=\sqrt{5}$

radius $r_2=\sqrt{g^2+f^2-c}=\sqrt{0+16+4}=2\sqrt{5}$

Now,

distance between the centers of the circles $c_1{c}_2=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$

$=\sqrt{(1-0{)}^2+(2-4{)}^2}=\sqrt{1+4}=\sqrt{5}$

We observe that

$r_1+r_2=\sqrt{5}+2\sqrt{5}=3\sqrt{5}$

$|r_1-r_2|=|\sqrt{5}-\sqrt[2]{25}|=|-\sqrt{5}|=\sqrt{5}$

So, $c_1{c}_2=|r_1-r_2|$

Distance between the circles is equal to the difference between the radiuses of the circles.

This is possible only when two circles touch each other internally.

We can also visualize it the above shown diagram.

Option A is correct

Key concept: 1. finding center of circle

2. Calculating radius of circle

3. Calculating distance between two circles