The straight line $2 x+3 y-k=0, k>0$ cuts the $x$ and $y$-axes at $\mathrm{A}$ and $\mathrm{B}$. The area of $\triangle \mathrm{OAB}$, where $\mathrm{O}$ is the origin, is 12 sq unit. The equation of the circle having $A B$ as diameter is

• Correct answer
  A

  $x^{2}+y^{2}-6 x-4 y=0$
• B

  $x^{2}+y^{2}+4 x-6 y=0$
• C

  $x^{2}+y^{2}-6 x+4 y=0$
• D

  $x^{2}+y^{2}-4 x-6 y=0$

Solution

Given, equation of line, $2 x+3 y-k=0, k>0$. $\frac{x}{\frac{k}{2}}+\frac{y}{\frac{k}{3}}=1 \quad \text{...(i)}$

Given area of $\triangle \mathrm{OAB}=12$ $$ \begin{gathered} =\frac{1}{2}\left|\begin{array}{ccc} 0 & 0 & 1 \\ \frac{\mathrm{k}}{2} & 0 & 1 \\ 0 & \frac{\mathrm{k}}{3} & 1 \end{array}\right|=12 \\ \Rightarrow \quad \frac{\mathrm{k}^{2}}{6}=24 \Rightarrow \mathrm{k}^{2}=144 \Rightarrow \mathrm{k}=12 \end{gathered} $$ Hence, the coordinate of $A(6,0)$ and $B(0,4)$. Then, diameter $=\mathrm{AB}$ $$ \begin{aligned} &=\sqrt{(6-0)^{2}+(0-4)^{2}} \\ &=\sqrt{36+16} \\ &=\sqrt{52} \end{aligned} $$ Centre $=$ Mid point of $A B$ $$ =\left(\frac{6+0}{2}, \frac{0+4}{2}\right)=(3,2) $$ The equation of circle is $\Rightarrow \quad(x-3)^{2}+(y-2)^{2}=\left(\frac{\sqrt{52}}{2}\right)^{2}$ $\Rightarrow \mathrm{x}^{2}+9-6 \mathrm{x}+\mathrm{y}^{2}+4-4 \mathrm{y}=13$ $\Rightarrow \quad x^{2}+y^{2}-6 x-4 y+13=13$ $\Rightarrow \quad x^{2}+y^{2}-6 x-4 y=0$

Hence, the correct answer is option: 'A'.