How to find the standard form of the equation of the specified circle given that it Is tangent to line x+y =2 at point (4,-2) and the center is on the x-axis?

1 Answer

Answer:

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

Explanation:

Let #(x_1, 0)# be the center on the x-axis & #r# be the radius of circle then the equation of circle

#(x-x_1)^2+(y-0)^2=r^2#

#(x-x_1)^2+y^2=r^2#

Now, since the above circle is tangent to the line #x+y=2# at #(4, -2)# hence the point #(4, -2)# will satisfy the equation of circle

#(4-x_1)^2+(-2)^2=r^2#

#r^2=x_1^{2}-8x_1+20\ .....(1)#

Now, the line #x+y=2# is tangent to the circle hence, substituting #y=2-x# in equation of circle we get

#(x-x_1)^2+(2-x)^2=r^2#

#2x^2-2(x_1+2)x+x_1^{2}+4-r^2=0#

The above equation will have equal roots iff discriminant #B^2-4AC=0#

#(-2(x_1+2))^2-4(2)(x_1^{2}+4-r^2)=0#

#x_1^{2}-4x_1+4-2r^2=0#

substituting the value of #r^2# from (1) in above equation, we get

#x_1^{2}-4x_1+4-2(x_1^{2}-8x_1+20)=0#

#x_1^{2}-12x_1+36=0#

#(x_1-6)^2=0#

#x_1=6, 6#

we get sinhle real value of #x# i.e. #x=6#, substituting this value in (1), we get

#r^2=6^{2}-8\cdot 6+20#

#r^2=8#

hence, the equation od circle is

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

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