
Centre (1,−2),r=3 Reflection of (1,−2) about 2x−3y+5=0 $\begin{aligned}
& \frac{x-1}{2}=\frac{y+2}{-3}=\frac{-2(2+6+5)}{13}=-2 \
& x=-3, y=4
\end{aligned}Equationofcircle′C′C:(x+3)^2+(y-4)^2=9$ A.T.Q.
ℓ(arcAB)=61×2πrrθ=61×2πrθ=3π(α+6)2+(β−4)2=27(α+6)2−(α+3)2=18(α+3)2±(β−4)2=9⇒6α=−9 ⇒α=2−3,β=(4−233)∴β−3α(4−233)+233=4