Question

2 perpendicular chords AB and CD of a circle intersect at O (inside the circle). AB is bisected at O. If AO = 4 units and OD = 6 units. Find circumradius of $△ACO$.

• A. $\sqrt{13}$
• B. $\frac{2\sqrt{13}}{3}$
• C. $\frac{\sqrt{15}}{2}$
• D. $\frac{2\sqrt{15}}{4}$

Answer 1

Answer: (B)

$\frac{2\sqrt{13}}{3}$

$\angle BDO=\angle CAO=A$ as angles are made by a chord of circle on its surface are equal.

By $AA;\Delta DOB\sim \Delta AOC$

Therefore sides will be in equal proportion.

$\frac{OA}{OC}=\frac{OD}{OB}$

$⟹\frac{4}{OC}=\frac{6}{4}⟹OC=\frac{16}{6}=\frac{8}{3}$ units

$AC=\sqrt{4^2+{\left(\frac{8}{3}\right)}^2}=\sqrt{16+\frac{64}{9}=}\sqrt{\frac{144+64}{9}=}\sqrt{\frac{208}{9}}=\frac{4\sqrt{13}}{3}$ units

Therefore circumradius of $\Delta ACO=\frac{\left(OA\right).\left(OC\right).\left(AC\right)}{4\Delta }=\frac{4\times \frac{8}{3}\times \frac{4\sqrt{13}}{3}}{4\times \frac{1}{2}\times 4\times \frac{8}{3}}=\frac{2\sqrt{13}}{3}$

Hence,circumradius of $\Delta ACO=\frac{2\sqrt{13}}{3}$