Question 602433
If the line {{{mx-y=0}}} <--> {{{y=mx}}} is tangent to the circle, they have only one point in common. So, substituting
{{{y=mx}}} into the equation for the circle,
{{{x^2+y^2-6x+7y+2=0}}}, there should be only one point that is a solytion.
That would mean only one pair (x,y), meaning only one value for x.
So, here we go
{{{x^2+(mx)^2-6x+2(mx)+2=0}}} --> {{{x^2+m^2x^2-6x+2mx+2=0}}} -->  {{{(m^2+1)x^2+(2m-6)x+2=0}}}
If a quadratic equation {{{ax^2+bx+c=0}}} has only one solution its discriminant must be zero:
{{{b^2-4*a*c=0}}}
For the equation with m,
{{{a=m^2+1}}}, {{{b=2m-6}}} and {{{c=2}}}
The discriminant is
{{{(2m-6)^2-4*(m^2+1)*2=0}}} --> {{{4m^2-24m+36-8(m^2+1)=0}}} --> {{{4m^2-24m+36-8m^2-8=0}}} --> {{{-4m^2-24m+28=0}}}
Dividing both sides of the equal sign by (-4), we get as much friendlier equation:
{{{m^2+6m-7=0}}}
and factoring, we get
{{{(m+7)(m-1)=0}}} with solutions {{{highlight(m=1)}}} and {{{highlight(m=-7)}}} .
{{{drawing(300,300,-2,8,-7,3,
grid(1),
circle(3,-1,2.8284),
line(-2,-2,3,3), line(-1,7,1,-7)
)}}}