Name the points as shown. Let H be the intersection of AO and BC. Since \(\Delta ABC\) isosceles at A, AH is the perpendicular bisector of BC

\(\Rightarrow BH=\dfrac{BC}{2}=\dfrac{6}{2}=3\) (inches)

 \(\Delta AHB\) right at H has : \(AH=\sqrt{AB^2-BH^2}=\sqrt{5^2-3^2}=4\) (inches) (Pythagorean theorem)

Denote OA = OB = x (inches), then OH = 4 - x

\(\Delta OHB\) right at H has : \(9+\left(4-x\right)^2=x^2\) (Pythagorean theorem)

\(\Leftrightarrow9+16-8x+x^2=x^2\Leftrightarrow8x=25\Leftrightarrow x=3\dfrac{1}{8}\)

So, the radius of the circle is \(3\dfrac{1}{8}\) inches

Applying the Pyth