Question

Find the equation of circle touching the line x = 2 and x = 12 and passes through (4 , 5) ?

Answer 1

Equation is either `x^2+y^2-14x-18y+105=0` or `x^2+y^2-14x-2y+25=0`

Explanation:

As the circle touches both `x=2` as well as `x=12`,

its center lies on `x=(2+12)/2=7` i.e. it is of type `(7,k)`

and radius is `(12-2)/2=5`

Hence the equation could be

`(x-7)^2+(y-k)^2=25`

and as it passes through `(4,5)`

`(4-7)^2+(5-k)^2=25`

or `(5-k)^2=25-9=16`

i.e. `5-k=+-4` and `k=1` or `k=9`

and hence equation of circle is `(x-7)^2+(y-1)^2=25`

i.e. `x^2+y^2-14x-2y+25=0`

or `(x-7)^2+(y-9)^2=25`

i.e. `x^2+y^2-14x-18y+105=0`

graph{(x^2+y^2-14x-18y+105)(x^2+y^2-14x-2y+25)(x-2)(x-12)((x-4)^2+(y-5)^2-0.05)=0 [-17.59, 22.41, -5.84, 14.16]}