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?

1 Answer
Jul 4, 2016

Answer:

#=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#