Physics, Chemistry, Mathematics Forums

is the answer to the first problem a) 7
here is my solution

starting with k=1, the particle moves 1 unit.so when it stops on c₁ it has covered an angle of 1 (radian) on c₁ (angle=arc/radius = 1/1 =1)
now it moves radially to c₂. At this position on c₂ angle is still 1 but now radius is 2.So length of arc already covered by particle on c₂ is 2 (arc = angle*radius = 1*2 =2 )
now it moves 2 units on c₂ .So it has actually covered 2+2 =4 units on c₂ .So new angle subtended = 2 radian (arc/radius =4/2 =2)
now it moves radially to c₃. At this position on c₃ angle is still 2 but now radius is 3.So length of arc already covered by particle on c₃ is 6 (arc = angle*radius = 2*3 =6 ).then it moves 3 units.so actual distance covered = 6+3 = 9 units on c₃ .so new angle subtended = 3 radian.
go on like this till the angle subtended is 7 radian u'll get k=7.the particle crosses the +ve direction of x-axis when angle subtended is greater than 2pi radians(6.28 radians).

and the second question......

the centre of circle is O(p/2,q/2)
let the chords be bisected at P(a,0)
so OP is perpendicular to the chords
slope m_op = (q/2)/(p/2 - a)
               = q/(p-2a)

slope m_chords = -1/m_op = (2a-p)/q

Equation of chords passing through P(a,0) and having slope m_chords is
y - 0 = (2a-p)/q (x- a)
=> qy = (2a-p)(x-a)
this chord passes through (p,q)
therefore
q² = (2a-p)(p-a)
q² + p² = 3ap - 2a²
2a² - 3ap + p² + q² = 0
since 2 real distinct chords are bisected by x-axis you will get 2 real distinct values of a from the above quadratic equation
hence D>0
solve it and u'll get p² > 8q²