How do you write an equation for a circle with centre \[\left( 3,5 \right)\] tangent to the X-axis?

Hint: To solve the given problem, we should know the properties of a circle. 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 distance between the tangent and centre of a circle is equal to the radius.

Complete step by step answer:
We are asked to find the equation of the circle with centre \[\left( 3,5 \right)\] tangent to the X-axis. To do this, we will use the 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.
We are already given the coordinates of the centre as \[\left( 3,5 \right)\]. Thus, we have the values of h and k as 3 and 5 respectively. We also know that the X-axis is a tangent to the given circle. Using the property that the distance between the tangent and centre of a circle is equal to the radius. The radius of the circle equals distance between the point \[\left( 3,5 \right)\] and X-axis. The distance equals 5.
Now that we have both centre and radius, we can write the equation of the circle as \[{{\left( x-3 \right)}^{2}}+{{\left( y-5 \right)}^{2}}={{5}^{2}}\].
Using this equation, we can also graph it as



Note: We can use this question to form a property as follows.
The equation of a circle with centre at \[\left( h,k \right)\] and having X-axis as its tangent is \[{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{k}^{2}}\]. Here, \[\left( h,k \right)\] are the coordinates of the centre of the circle.