The radius of a circle with centre O is $$\sqrt{50}$$cm. A and C are two points on the circle, and B is a point inside the circle. The length of AB is 6 cm, and the length of BC is 2 cm. The angle ABC is a right angle. Find the square of the distance OB.

We know that ABC is a right angled triangle.
=> $$AC=\sqrt{6^2+2^2}$$
$$AC=\sqrt{40}$$
$$AC=2*\sqrt{10}$$
Let the coordinates of A be (0,0)
We know that the radius of the circle, OA = $$\sqrt{50}$$cm
Let OD be the height of the triangle AOC.
$$AC/2=*\sqrt{10}$$

By applying Pythagoras theorem, we get,
=>The height(y) of  the point  O $$= \sqrt{50-10}$$
The height(y) of  the point  O$$=\sqrt{40}$$cm
=> Coordinates of point O = $$(\sqrt{10},\sqrt{40})$$

Area of triangle ABC = $$0.5*6*2$$ = $$6$$ square units.
Let the height of triangle ABC be h.
0.5*h*AC=6 
h*AC = 12
h*$$2*\sqrt{10}$$ = $$12$$
h = $$\frac{6}{\sqrt{10}}$$
X-coordinate of point B =$$\sqrt{6^2-\frac{36}{10}}$$
                                    = $$\sqrt{32.4}$$ cm
Distance between points O and B = $$\sqrt{(\sqrt{10}-\sqrt{32.4})^2+({\sqrt{40}-\frac{6}{\sqrt{10}})^2}}$$
Expanding, we get,
Square of the distance between points O and B = $$26$$ cm.
Therefore, option A is the right answer.