Question 1184256




{{{x^2+y^2=10x}}}....eq.1

{{{y=kx-10 }}}.....eq.2
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substitute {{{y}}} in eq.1

{{{x^2+(kx-10)^2=10x}}}

{{{x^2+k^2x^2 - 20kx -10x+ 100=0}}}

{{{(1+k^2)x^2 - (20k +10)x+ 100=0}}}


if the line meets the curve, there is one solution and it is a case if discriminant ={{{0}}}


so, use discriminant 

{{{b^2-4ac=0}}}  ...in your case {{{a=(1+k^2)}}}, {{{b=- (20k +10)}}}, and {{{c=100}}}

{{{(- (20k +10))^2-4(1+k^2)*100=0  }}}

{{{400k^2 + 400k + 100-400 k^2 - 400=0 }}}
 
 {{{400k - 300=0  }}}

{{{ 400k = 300 }}}
 
{{{4k = 3}}}

{{{k = 3/4}}}

and your line is:

{{{y=(3/4)x-10 }}}



{{{ graph( 600, 600, -10, 10, -10, 10, (3/4)x-10, sqrt(10x-x^2), -sqrt(10x-x^2)) }}}