Question 22

Let circle-1 and circle-2 are orthogonal circles. The length of an arc measuring $$60^\circ$$ of circle-1 is equal to the length of an arc measuring $$45^\circ$$ of circle-2. Then the ratio of the area of square on the line-segment joining their centres to the area of square on the radius of circle-1, in its simplest form, is

[Note: Let two distinct circles are intersecting in different points. If their tangents at the point of intersection are perpendicular to each other, then the circles are said to be Orthogonal circles]

Correct Answer: $$\large \frac{25}{9}$$

Solution

Equal arc lengths give $$\frac{60}{360} \times 2\pi r_1 = \frac{45}{360} \times 2\pi r_2$$, so $$r_2 = \frac{4}{3} r_1$$. For orthogonal circles the tangents at a common point are perpendicular, so the radii there are perpendicular and $$d^2 = r_1^2 + r_2^2 = r_1^2 + \frac{16}{9}r_1^2 = \frac{25}{9}r_1^2$$. The required ratio of the two squares is $$\frac{d^2}{r_1^2} = \frac{25}{9}$$, that is $$25 \colon 9$$, which is about $$2.78$$.