Given that EF and EG are tangent lines, apply the Tangent Segments Theorem to set up an equation and solve for x.

Answer:The value of x is 3Step-by-step explanation:we know thatThe Tangent Segment Theorem states that if from one external point, two tangents are drawn to a circle then they have equal tangent segmentsso In this problem[tex]EF=EG[/tex]substitute the values[tex]x^{2}-4x+5=2x-4\\x^{2}-4x+5-2x+4=0\\x^{2}-6x+9=0[/tex]Solve the quadratic equationThe formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to [tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex] in this problem we have [tex]x^{2}-6x+9=0[/tex]  so [tex]a=1\\b=-6\\c=9[/tex] substitute in the formula [tex]x=\frac{6(+/-)\sqrt{-6^{2}-4(1)(9)}} {2(1)}[/tex] [tex]x=\frac{6(+/-)\sqrt{0}} {2}[/tex] [tex]x=\frac{6} {2}=3[/tex]