Circles are drawn through the points \[\left( {2,0} \right)\]to cut the intercept of length \[5\] units on the x-axis. If their center lies in the first quadrant, then find their equation.
A. \[{x^2} + {y^2} - 9x + 2ky + 14 = 0\]
B. \[3{x^2} + 3{y^2} + 27x - 2ky + 42 = 0\]
C. \[{x^2} + {y^2} - 9x - 2ky + 14 = 0\]
D. \[{x^2} + {y^2} - 2kx - 9y + 14 = 0\]

Hint:
In order to solve the question, first find the center of the circle. Then, find the radius of the circle. Finally, using the general form of the equation of the circle, find the required equation.



Formula Used:
Equation of the circle is \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]
\[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]



Complete step-by-step answer:
Consider the figure shown below.

Image: The diagram represents the circle
Here the length of AB is 5 units. So the center is
\[ = \left( {\dfrac{{2 + 7}}{2},k} \right)\]
\[ = \left( {\dfrac{9}{2},k} \right)\]
Also, here the radius is
\[r = \sqrt {{{\left( {\dfrac{5}{2}} \right)}^2} + {k^2}} \]
\[r = \sqrt {\dfrac{{25}}{4} + {k^2}} \]
We know that the equation of a circle is given as
\[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]
Where (h, k) is the center and r is the radius.
That is, we can write the equation as
                \[{\left( {x - \dfrac{9}{2}} \right)^2} + {\left( {y - k} \right)^2} = \dfrac{{25}}{4} + {k^2}\]
\[{x^2} + \dfrac{{81}}{4} - 9x + {y^2} + {k^2} - 2ky = \dfrac{{25}}{4} + {k^2}\] . . . . . . ( \[ \because {\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\])
          \[{x^2} - 9x + {y^2} - 2ky + \dfrac{{56}}{4} = 0\]
            \[{x^2} + {y^2} - 9x - 2ky + 14 = 0\]
Hence option C is the correct answer.



Note:
Students can get confused while taking the center points. Always remember that the center of a circle is the fixed point in the middle of the circle. Once the coordinates of the center and the radius are found the equation of the circle can be found easily.