Question

Area of part of circle x² + y² – 4x – 6y +12 = 0 above the line 4x + 7y – 29 = 0 is △, then [△] = ________ [.] is greatest integer function.

Answer 1

Since line 4x + 7y – 29 = 0

passes through the centre (2, 3) of the circle

∴ The line is a diameter of a circle with radius \(\sqrt{4 + 9 -12} = 1\) 

∴ area of semi circle is = \(\frac 12\pi r^2\)

\(=\frac 12\pi \)

\(= \frac{3.14}2\)

\(= 1.57\)

Hence [△] = [1.57] = 1