Q87 / 160 · AP EAMCET 2022 (07 Jul Shift 2) · Mathematics

The straight line touching the circle $x^2+y^2-2 x-3=0$ and remaining normal to the circle $x^2+y^2-4 y-6=0$ is

  • Correct answer
    A
    $4 x-3 y+6=0$
  • B
    $y+2=0$
  • C
    $4 x+3 y-6=0$
  • D
    $2 x+3=0$

Solution

Line is normal to circle $x^2+y^2-4 y-6=0$ $\therefore$ It passes through centre $(0,2)$. Let $m$ be slope of line. $\therefore$ Equation of line is $y-2=m x$ $$ \Rightarrow \quad m x-y+2=0 $$ This line is tangent to $x^2+y^2-2 x-3=0$ centre is $(1,0)$, radius is 2 $$ \begin{aligned} & \therefore \quad 2=\left|\frac{m+2}{\sqrt{1+m^2}}\right| \\ & \Rightarrow \quad 4\left(1+m^2\right)=(m+2)^2 \\ & \therefore \quad 3 m^2-4 m=0 \\ & \Rightarrow \quad m=0 \\ & \text { or } \quad m=\frac{4}{3} \\ & \end{aligned} $$ $\therefore \quad$ Line can be $y=2$ or $\quad 4 x-3 y+6=0$

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

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