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

Hint: To solve the given problem we should know the following properties of circle and straight line. First thing we should know is the equation of circle in centre-radius form which is expressed as \[{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}\]. Here, \[\left( h,k \right)\] are the coordinates of the centre of the circle, and r is the radius of it. Next, we should know that the slope of normal to the line with slope m is \[\dfrac{-1}{m}\].

Complete step by step answer:
We are asked to write the equation of the circle with the given features. As the coordinates of the centre lie on X-axis, let’s assume the centre at \[(a,0)\], and the radius of the circle as r. using this, the centre-radius form of the circle can be expressed as \[{{\left( x-a \right)}^{2}}+{{\left( y \right)}^{2}}={{r}^{2}}\].
We still have two unknown constants in this equation, to find their value we will use the other information given. As \[\left( 4,2 \right)\] is the point of contact of the tangent and circle, it must lie on the circle. Substituting this point in the equation of the circle, we get \[{{\left( 4-a \right)}^{2}}+{{\left( 2 \right)}^{2}}={{r}^{2}}\]. Also, the line joining the centre and the point of contact must be normal to the given tangent \[x-y=2\]. The slope of the tangent is 1. The slope of this normal using the two points can be calculated as \[\dfrac{2-0}{4-a}\]. We can also calculate the slope of the normal using tangent slope as \[\dfrac{-1}{1}\].
These two values of slope must be the same. Thus, \[\dfrac{2-0}{4-a}=\dfrac{-1}{1}\]. Solving this equation, we get the value of an equals to 6.
We can find the other variable using this value and the equation \[{{\left( 4-a \right)}^{2}}+{{\left( 2 \right)}^{2}}={{r}^{2}}\]. By doing this, we get \[{{r}^{2}}=8\].
Now, we have both the centre and radius of the circle. We can write the equation as \[{{\left( x-6 \right)}^{2}}+{{\left( y \right)}^{2}}=8\].
We can also plot the circle as follows:


Note: To solve these types of questions, we should know the properties of the circle and tangents of the circles. The radius of the circle is perpendicular at the point of contact of tangent and circle, and the distance between the centre and point of contact of tangent is equal to the radius of the circle.