Question
Consider the two curves $${C_1}:{y^2} = 4x,\,{C_2}:{x^2} + {y^2} - 6x + 1 = 0.$$ Then,
A.
$${C_1}\,{\text{and}}\,{C_2}$$ touch each other only at one point.
B.
$${C_1}\,{\text{and}}\,{C_2}$$ touch each other exactly at two points
C.
$${C_1}\,{\text{and}}\,{C_2}$$ intersect (but do not touch) at exactly two points
D.
$${C_1}\,{\text{and}}\,{C_2}$$ neither intersect nor touch each other
Answer :
$${C_1}\,{\text{and}}\,{C_2}$$ touch each other exactly at two points
Solution :
The given curves are
$${C_1}:{y^2} = 4x\,......\left( 1 \right)\,{\text{and}}\,{C_2}:{x^2} + {y^2} - 6x + 1 = 0\,......\left( 2 \right)$$
Solving (1) and (2) we get
$$\eqalign{
& {x^2} + 4x - 6x + 1 = 0 \Rightarrow x = 1\,{\text{and}}\, \Rightarrow y = 2\,{\text{or}}\, - 2 \cr
& \therefore {\text{Points of intersection of the two curves are}}\,\left( {1,2} \right)\,{\text{and}}\,\left( {1, - 2} \right) \cr
& {\text{For}}\,{{\text{C}}_{\text{1}}},\frac{{dy}}{{dx}} = \frac{2}{y} \cr
& \therefore {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1,2} \right)}} = 1 = {m_1} \cr
& {\text{and}}\,{\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1, - 2} \right)}} = - 1 = {m_1}' \cr
& {\text{For}}\,{{\text{C}}_{\text{2}}},\frac{{dy}}{{dx}} = \frac{{3 - x}}{y}\,\therefore {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1, - 2} \right)}} = 1 = {m_2} \cr
& {\text{and}}\,{\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1, - 2} \right)}} = - 1 = {m_2}' \cr
& \because {m_1} = {m_2}\,{\text{and}}\,{m_1}' = {m_2}' \cr
& \therefore {C_1}\,{\text{and}}\,{C_2}\,{\text{touch each other at two points}}{\text{.}} \cr} $$