Question

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

• A. $x^2+y^2-14x-6y+49=0$
• B. $x^2+y^2-14x-6y+9=0$
• C. $x^2+y^2-6x-14y+49=0$
• D. $x^2+y^2-6x-14y+9=0$

Answer 1

Answer: (A)

$x^2+y^2-14x-6y+49=0$

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

The coordinates of $C$ are $(OL,LC)$ $OL=OB+BL=3+5\sin \theta$ $=3+5\times \left(\frac{4}{5}\right)=7$ $CL=5\cos \theta =5\times \left(\frac{3}{5}\right)=3$
So, the coordinate of $C$ are $(7,3)$ and the equation of the circle having $C$ as centre and touching $x$-axis is
$(x-7{)}^2+(y-3{)}^2=(CL{)}^2=9$
$\Rightarrow x^2+y^2-14x-6y+49=0$