Circle drawn through the point $(2,0)$ to cut intercept of length $‘5^{\prime }$ units on the x-axis. If its centre lie in the first quadrant then the equation of family of such circles is

The correct option is B. $x^2+y^2-9x-2ky+14=0,k>0$

$AB=5$
Centre $=(\frac{2+7}{2},K)$
$=(\frac{9}{2},K)$
Radius $=\sqrt{\frac{25}{4}+K^2}$
Equation of circle is
${(x-\frac{9}{2})}^2+(y-k{)}^2=\frac{25}{4}+k^2$
$\Rightarrow x^2+\frac{81}{4}-9x+y^2+k^2-2ky=\frac{25}{4}+k^2$
$\Rightarrow x^2-9x+y^2-2ky+\frac{56}{4}=0$
$\therefore x^2+y^2-9x-2ky+14=0$