Answer:
The value of x is 3
Step-by-step explanation:
we know that
The Tangent Segment Theorem states that if from one external point, two tangents are drawn to a circle then they have equal tangent segments
so
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 equation
The 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]
