Two Perpindicular Curves

Let $O$ be the centre of the circle. Drop perpendiculars from the centre to the chords $AB$ and $CD$ intersecting the chords at $P$ and $Q$ respectively. Since, all the angles of quadrilateral $OPEQ$ are 90 degrees, hence we can conclude that $OPEQ$ is a rectangle with $OP=EQ$ and $PE=OQ$ . Length of chord $AB=BE+EA=5+2=7$ Hence, $BP=PA=3.5$ (because perpendicular from the centre to a chord bisects the chord) Hence, $PE=BE-BP=5-3.5=1.5$ .

Let $EQ=x$ , and the radius of the circle be $r$ . Using pythagoras theorem in triangle OQC, $r^2=(1.5)^2+(x+3)^2$ . Using pythagoras theorem in triangle OPB, $r^2=(3.5)^2+x^2$ . Solving, we get $x=1/6$ and $r= \sqrt{442} /6$ hence, $a+b=442+6=448$ .

[LaTex Edits - Calvin]

By Power of a Point, we have $ED=\frac{10}{3}$ . Now drop the perpendiculars from the center $O$ to $AB$ and $CD$ at $F$ and $G$ . These bisect the chords, so we have $BF=\frac{7}{2}$ and $DG=\frac{19}{6}$ . Therefore, $EG=\frac{1}{6}$ and by the Pythagorean theorem we have $r^2=BF^2+EG^2=\frac{442}{36} \rightarrow r=\frac{\sqrt{442}}{6}$ and our desired answer is $448$ .

First we find ED by using BE * EA = EC * ED. ED = 10/3

Then we join B,D and D,A forming the triangle BDA.

Now, as the chords intersect at right angles, triangle BED , DEC are all right angled triangles. WE find the values of BD and AD using pythagorus's theorem. Incidentally, BD = root325 / 3 and AD = root136/3

Then we find the area of triangle BAD, taking DE as the altitude and AB as the base, which comes to 35/3.

Finally, we use the well known relation , abc/4R = area of the triangle, where a,b,c are the three sides of the triangle.

Plugging the values into this equation and doing some algebra, we finally reach root442/6 and hence the answer.

sorry gor not writing in latex, some keys of my keyboard dont work.. :/

ED= 2*5/3=10/3 (since AE X EB = CE X ED)

We also know that $AE^{2} + EB^{2} + EC^{2} + ED^{2} = 4r^{2}$

which gives $4+25+9+\frac{100}{9}=\frac{442}{9}=4r^{2}$

Thus $r=\frac{\sqrt{442}}{6}$