Q95 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics
The equation of one of the common tangents of the circle $x^2+y^2-6 y+4=0$ and the parabola $y^2=x$ is
- A$2 x-y+1=0$
- B$2 x-y=1$
- C$4 x-y+1=0$
- Correct answerD$x-2 y+1=0$
Solution
Let the equation of tangent to the parabola $y^2=x$, having slope $m$ is,
$$
y=m x+\frac{1}{4 m}
$$
For common tangent to the circle
$$
x^2+y^2-6 y+4=0
$$
$\therefore$ Radius $\sqrt{9-4}=\frac{\left|3-\frac{1}{4 m}\right|}{\sqrt{1+m^2}}$
$$
\begin{array}{rlrl}
\Rightarrow & 5\left(1+m^2\right) & =9+\frac{1}{16 m^2}-\frac{3}{2 m} \\
\Rightarrow & & m & =\frac{1}{2}
\end{array}
$$
So, equation of common tangent is
$$
\begin{aligned}
y & =\frac{1}{2} x+\frac{1}{2} \\
\Rightarrow \quad x-2 y+1 & =0 .
\end{aligned}
$$
Hence, the correct answer is option: 'D'.