Question

A chord AB of a circle is at a distance 8cm less than the radius from the centre. If radius of a circle is less than chord by 11cm then find the length of chord, radius and the distance of chord from the centre.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{radius}=\mathrm{OA}=\mathrm{OB}=x\mathrm{cm} \\ \mathrm{Then}\mathrm{OM}=(x-8)\mathrm{cm} \\ \mathrm{And}\mathrm{AB}=(x+11)\mathrm{cm} \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{perpendicular}\mathrm{from}\text{the}\mathrm{centre}\mathrm{on}a\mathrm{chord}\mathrm{bisects}\mathrm{the}\mathrm{chord}. \\ \therefore \mathrm{AM}=\frac{\mathrm{AB}}{2}=\left({\displaystyle \frac{x+11}{2}}\right)\mathrm{cm} \\ \mathrm{Now},\mathrm{in}\mathrm{right}△\mathrm{OAM}, \\ {\mathrm{OA}}^2={\mathrm{OM}}^2+{\mathrm{AM}}^2 \\ \Rightarrow x^2={\left(x-8\right)}^2+{\left({\displaystyle \frac{x+11}{2}}\right)}^2 \\ \Rightarrow x^2={\left(x-8\right)}^2+\frac{{\left(x+11\right)}^2}{4} \\ \Rightarrow 4x^2=4{\left(x-8\right)}^2+{\left(x+11\right)}^2 \\ \Rightarrow 4x^2=4x^2+256-64x+x^2+22x+121 \\ \Rightarrow x^2-42x+377=0 \\ \Rightarrow x^2-13x-29x+377=0 \\ \Rightarrow x\left(x-13\right)-29\left(x-13\right)=0 \\ \Rightarrow \left(x-13\right)\left(x-29\right)=0 \\ \Rightarrow x=13\mathrm{or}x=29 \\ \mathrm{When}x=13, \\ \mathrm{chord}=x+11=24\mathrm{cm} \\ \mathrm{radius}=13\mathrm{cm} \\ d\mathrm{istance}\mathrm{of}\mathrm{chord}\mathrm{from}\text{the}\mathrm{centre}=x-8=5\mathrm{cm} \\ \mathrm{When}x=29, \\ \mathrm{chord}=x+11=40\mathrm{cm} \\ \mathrm{radius}=29\mathrm{cm} \\ \mathrm{distance}\mathrm{of}\mathrm{chord}\mathrm{from}\text{the}\mathrm{centre}=x-8=21\mathrm{cm} \end{gathered}$$