On the line joining the points $A (0,4)$ and $B (3, 0)$, a square $ABCD$ is constructed on the side of the line away from the origin. Equation of the circle having centre at $C$ and touching the axis of $x$ is

Let $\angle ABO = \theta$, then $\angle CBL= 90^{\circ} -\theta, CL$ being perpendicular to x-axis $(Fig)$.

The coordinates of $C$ are $(OL, LC)$ $OL = OB + BL= 3 + 5 \sin \theta$ $= 3 + 5 \times \left (\dfrac 45\right) = 7$ $CL = 5 \cos \theta = 5 \times \left (\dfrac 35\right) =3$

So, the coordinate of $C$ are $(7, 3)$ and the equation of the circle having $C$ as centre and touching $x$-axis is

$(x -7)^{2} + (y-3)^{2} = (CL)^{2} = 9$

$\Rightarrow x^{2} + y^{2} -14x -6y + 49 = 0$