Question

Show that the circle x² + y² -2x -4y=0 and x² + y² - 8x -12y +48=0 touches externally and find the point of contact?

Answer 1

Refer to exlpanation

Explanation:

The first circle is written as follows

`x^2+y^2-2x-4y=0=>(x^2-2x+1)+y^2-4y+4)=(sqrt5)^2=>(x-1)^2+(y-2)^2=(sqrt5)^2`

The second circle is written as follows

`x² + y² - 8x -12y +48=0=>(x^2-8x+16)+(y^2-12y+36)=52-48=>(x-4)^2+(y-6)^2=(2)^2`

The centre of first circle is point `A(1,2))` and the centre of second centre is point `B(4,6)` and their distance is

`D=sqrt((4-1)^2+(6-2)^2)=sqrt(9+16)=sqrt25=5`

The distance of the sum of their radius is `2+sqrt5=4.23` which is smaller than the distance of their centre.Hence they dont touch externally.

Answer 2

For the modified question in Shafiyah's comment..

Explanation:

`x^2+y^2-2x-4y-4=0`

and

`x^2+y^2-8x-12y+48=0`

The first circle:

`x^2-2x " " " " +y^2-4y " " " " =4`

Complete the two squares by adding the squares of half the coefficient of the linear terms.

`x^2-2x underbrace(+1)+y^2-4y underbrace(+ 4) = 4 +1 +4`

`(x-1)^2 + (y-2)^2 = 9`

Is a circle with center `(1,2)` and radius `3`.

The second circle

`x^2-8x " "" "+y^2-12y " "" "=-48`

`x^2-8x underbrace(+16) +y^2-12y" underbrace(+36) = -48+16+36`

`(x-4)^2 + (y-6)^2 = 4`

Is a circle with center `(4,6)` and radius `2`.

The distance between the centers is equal to the sum of the radii, so the circles touch at one point.

There are several ways of finding the point of intersection.

It is `(14/5,22/5) = (2.8, 4.4)`