Question 985482


To find the intersection points of the circle and the straight line,  substitute  {{{y}}} = {{{mx + 4}}}  into the equation  {{{x^2 + y^2}}} = {{{4}}}.  You will get

{{{x^2}}} + {{{(mx+4)^2}}} = {{{4}}},


{{{x^2}}} + {{{m^2*x^2 + 2*4*mx + 4}}} = {{{4}}},


{{{(1 + m^2)*x^2 + 8mx + 16}}} = {{{4}}},


{{{(1 + m^2)*x^2 + 8mx + 12}}} = {{{0}}}.


Next,  apply the quadratic formula to find the roots


{{{x}}} = {{{(-8m +- sqrt(d))/(2*(1+m^2))}}},


where  {{{d}}}  is the discriminant  {{{d}}} = {{{64*m^2 - 4*12*(1+m^2)}}} = {{{(64-48)*m^2 - 48}}} = {{{16*m^2 - 48}}}. 


Now everything is determined by the discriminant. 


If  {{{d}}} > {{{0}}},  then there are two roots.  Correspondingly,  there are two intersection points. 

In opposite,  if there are two intersection points,  then there are two roots, hence, {{{d}}} > {{{0}}}.


If  {{{d}}} = {{{0}}},  then there is only one root.  It means that the straight line tangents the circle. 

And in opposite,  if the straight line tangents the circle, then there is only one root,  hence,  {{{d}}} = {{{0}}}.


Finally, &nbsp;{{{d}}} < {{{0}}} <-----> there are no intersection points. 


Thus you need to solve this critical equation


{{{16*m^2 - 48}}} = {{{0}}}.


It gives


{{{m}}} = +/- {{{sqrt(3)}}}. 


So, &nbsp;if &nbsp;{{{m}}} = +/- {{{sqrt(3)}}}, &nbsp;then the straight line tangents the circle.


If &nbsp;{{{abs(m)}}} > {{{sqrt(3)}}}, &nbsp;then &nbsp;{{{d}}}>{{{0}}}&nbsp; and there are two intersection points. 


If &nbsp;{{{-sqrt(3)}}} < {{{m}}} < {{{sqrt(3)}}}, &nbsp;then {{{d}}}<{{{0}}}&nbsp; and there are no intersection points.