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
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
$2 x-y+1=0$
$2 x-y=1$
$4 x-y+1=0$
$x-2 y+1=0$
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}
$