Question

The parabola $y^2=4x$ and the circle $x^2+y^2-6x+1=0$ will

• A. Intersect exactly at one point
• B. Touch each other at two distinct points
• C. Touch each other at exactly one point
• D. Intersect at two distinct points

Answer 1

The correct option is D

Intersect at two distinct points

Explanation for correct option

The given equations are

$y^2=4x$ …..$\left(i\right)$

$x^2+y^2-6x+1=0$ ..…$\left(ii\right)$

Using $\left(i\right)$ in equation $\left(ii\right)$ we have

$$\begin{gathered} \Rightarrow x^2+4x-6x+1=0 \\ \Rightarrow x^2-2x+1=0 \\ \Rightarrow {\left(x-1\right)}^2=0 \\ \therefore x=1 \end{gathered}$$

Now putting the value of $x$ in equation $\left(i\right)$ we have

$$\begin{gathered} \Rightarrow y^2=4 \\ \Rightarrow y=\pm 2 \end{gathered}$$

Therefore they intersects at two distinct points and the points are $\left(1,2\right)$ and $\left(1,-2\right)$

Hence, the correct option is $\left(\mathrm{D}\right)$.