Answer: The radius of the circle o is 8.43 unit.
Explanation:
It is given that AB is tangent to circle O at B, AB = 5 unit and AO = 9.8 unit.
According to the tangent perpendicular to radius theorem the tangent line and radius are perpendicular to the point of tangency.
Since the center of the circle is O, point of tangency is B and AB is the tangent line, therefore the line OB is perpendicular to AB. So the triangle ABO is a right angle triangle.
Use pythagoras theorem,
[tex](hypotenuse)^2=(base)^2+(perpendicular)^2[/tex]
[tex]AO^2=AB^2+OB^2[/tex]
[tex](9.8)^2=r^2+(5)^2[/tex]
[tex]96.04=r^2+25[/tex]
[tex]71.04=r^2[/tex]
[tex]r=\sqrt{71.04}[/tex]
[tex]r=8.428\approx 8.43[/tex]
Therefore, the radius of the circle o is 8.43 unit.
