Question

Consider the two curves $C_1:y^2=4x,C_2:x^2+y^2-6x+1=0.$ Then,

• A. $C_1$ and $C_2$ touch each other only at one point
• B. $C_1$ and $C_2$ touch each other exactly at two points
• C. $C_1$ and $C_2$ intersect (but do not touch) at exactly two points
• D. $C_1$ and $C_2$ neither intersect nor touch each other

Answer 1

Answer: (B)

The given curves, are $C_1:y^2=4x.....\left(1\right)$ and $C_2:x^2+y^2-6x+1=\mathrm{0.....}\left(2\right)$
Solving (1) and (2) we get
$x^2+4x-6x+1=0\Rightarrow x=1$ and $\Rightarrow y=2$ or $-2$
$\therefore$ Points of intersection of the two curves are $\left(1,2\right)$ and $\left(1,-2\right)$
For $C_1,\frac{dy}{dx}=\frac{2}{y}$
$\therefore {\left(\frac{dy}{dx}\right)}_{\left(1,2\right)}=1=m_1$ and ${\left(\frac{dy}{dx}\right)}_{\left(1,-2\right)}=-1=m_1^{\prime }$
For $C_2,\frac{dy}{dx}=\frac{3-x}{y}\therefore {\left(\frac{dy}{dx}\right)}_{\left(1,2\right)}=1=m_2$
and ${\left(\frac{dy}{dx}\right)}_{\left(1,-2\right)}=-1=m_2^{\prime }$
$\because m_1=m_2$ and $m_1^{\prime }=m_2^{\prime }$
$\therefore C_1$ and $C_2$ touch each other at two points.