Answer:
[tex]6\sqrt2[/tex]
Step-by-step explanation:
[tex]\text{Solution: }\\[/tex]
[tex]\text{Here, PF = PB = 6 [Radii of same circle are equal.]}[/tex]
[tex]\text{ME}=\dfrac{1}{2}\text{CB}=\dfrac{1}{2}\times6=3\ \ \ [\text{Radius of a circle is half of the diameter.}][/tex]
[tex]\text{Construction: }\text{Join P, F and M, E.}[/tex]
[tex]\text{1. In triangles PFD and MED,}\\\\\text{i. }\angle\text{PFD}=\angle\text{MED}=90^\circ\ \ \ (A)\ \ \ [\text{PF}\perp\text{ED}\text{ and }\text{ME}\perp\text{ED}, \text{radius is }[/tex]
[tex]\text{perpendicular to tangent at the point of}[/tex]
[tex]\text{contact.]}[/tex]
[tex]\text{ii. }\angle\text{PDE}=\angle\text{MDE}\ \ \ (A) \ \ \ [\text{Common angle.}][/tex]
[tex]\text{iii. }\triangle\text{PED}\sim\triangle\text{MED}\ \ \ [\text{By A.A. axiom.}][/tex]
[tex]\text{2. The corresponding sides of similar triangles are proportional, therefore,}\\\\\text{i. }\dfrac{\text{PF}}{\text{ME}}=\dfrac{\text{PD}}{\text{MD}}\\\\\\\text{or, }\dfrac{6}{3}=\dfrac{12+\text{CD}}{3+\text{CD}}[/tex]
[tex]\text{or, }2=\dfrac{12+\text{CD}}{3+\text{CD}}[/tex]
[tex]\text{or, }6+2\text{CD}=12+\text{CD}\\[/tex]
[tex]\text{or, CD}=6[/tex]
[tex]\text{3. Using pythagoras theorem in triangle MED,}\\\\\text{MD}^2=\text{ED}^2+\text{ME}^2\\\\\text{or, }(3+\text{CD})^2=\text{ED}^2+3^2\\\\\text{or, }(3+6)^2=\text{ED}^2+3^2\\\\\text{or, }81=\text{ED}^2+9\\\\\text{or, }\text{ED}^2=72\\\\\text{or, }\text{ED}=6\sqrt{2}[/tex]
