Question

How do you write an equation for a circle tangent to the line x - y = 2 at the point (4,2) and the center is on the x-axis?

Answer 1

`=x^2+y^2-12x+28=0`

Explanation:

Let the coordinate of the center of the circle lying on x-axis be `(a,0)` and its radius is r. The equation of the circle will be

`(x-a)^2+(y-0)^2=r^2`

`(x-a)^2+y^2=r^2.....(1)`

Now the point (4,2) is lying on the circle.So

`(4-a)^2+2^2=r^2`

`a^2-8a+20=r^2.....(2)`

Now it is given that `x-y=2` is the equation of tangent to the circle at the point(4,2) on the circle.
Witing the equation of the tangent in `y=mx +c` form we have the equation of the tangent as `y=x-2`,So it is obvious that the slope of the tangent is 1.
Hence the slope of the normal passing through (4.2) is `-1`

So equation of the normal at (4,2) will be

`(y-2)=(-1)(x-4)`

`=>y-2=-x+4`

`=>x+y=6....(3)`

Now as the center (a,0) is lying on the normal ,it will satisfy the equation of normal.
So inserting`x=a and y=0` in (3) we get
`a+0=6=>a=6`

Putting this value of a =6 in (2) we get

`a^2-8a+20=r^2`

`6^2-8xx6+20=r^2`

`=>r^2=8`

Now finally plugging in the value of `a =6 and r^2=8` in equation(1) we get the required equation of circle

`(x-6)^2+y^2=8`

`=>x^2-12x+36+y^2=8`

`=x^2+y^2-12x+28=0`