Q95 / 160 · AP EAMCET 2004 · Mathematics

If the line $3 x-2 y+6=0$ meets $X$-axis and $Y$-axis, respectively at $A$ and $B$, then the equation of the circle with radius $A B$ and centre at $A$ is

  • A
    $x^2+y^2+4 x+9=0$
  • Correct answer
    B
    $x^2+y^2+4 x-9=0$
  • C
    $x^2+y^2+4 x+4=0$
  • D
    $x^2+y^2+4 x-4=0$

Solution

Since, the line $3 x-2 y+6=0$ meets the coordinates axes, therefore $\therefore$ Points are $A(-2,0)$ and $B(0,3)$. $$ \because \text { Radius }=A B=\sqrt{2^2+3^2}=\sqrt{13} $$ $\therefore$ The equation of circle, whose centre $(-2,0)$ and radius $\sqrt{13}$, is $$ \begin{aligned} & (x+2)^2+(y-0)^2=(\sqrt{13})^2 \\ & \Rightarrow \quad x^2+y^2+4 x+4=13 \\ & \Rightarrow \quad x^2+y^2+4 x-9=0 \\ & \end{aligned} $$

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

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