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

1 Answer
Mar 5, 2018

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]}