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

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