Question

For all values of $\theta$ , the line $(x-2)cos\theta +(y-2)sin\theta =1$ touches a circle S = 0 whose centre is C and radius is R. A square is inscribed in the circle S = 0 such that its diagonals are parallel to the co - ordinate axes. Then

• A. Length of tangent from origin to the circle S = 0 is $\sqrt{7}$
• B. The vertices of the square are (1, 2), (2, 1), (3, 2), (2, 3).
• C. Area inside the circle but outside the square is $\pi -2$ sq. units
• D. Equation of one of the sides of the square is x - y - 1 = 0

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Length of tangent from origin to the circle S = 0 is $\sqrt{7}$
B The vertices of the square are (1, 2), (2, 1), (3, 2), (2, 3).
C Area inside the circle but outside the square is $\pi -2$ sq. units
D Equation of one of the sides of the square is x - y - 1 = 0

Equ. Of the circle is $(x-2{)}^2+(y-2{)}^2=1$
$\Rightarrow$ C = (2, 2), R = 1
(a) L.O.T from origin $=\sqrt{O{C}^2-R^2}=\sqrt{7}$
(b) C = (2, 2)
$\therefore$ Vertices =(2$\pm$ 1,2),(2,2$\pm$ 1)
= (3, 2), (1, 2), (2, 3), (2, 1)
(c) Required Area $=\pi (1{)}^2-(\sqrt{2}{)}^2=\pi -2$
(d) Equation of the side joining (2, 1), (3, 2) is x - y - 1 = 0.