Eqn of circle \((x-a)^2+(y-b)^2=r^2 => (x-1)^2+(y)^2=5^2\)
eqn of line y=mx since line is passing through (0,0) =>putting the values (4,8) we get \(y=2x\)
solving these equations we will get point A and B
(12/5,24/5) and (-2,-4)
midpoints will be for X \((12/5+(-2))/2=1/5\) and for Y \((24/5-4)/2=2/5\)

points are(1/5,2/5) sum= 3/5
C

\(? = w + z = {x_M} + {y_M}\,\,\,\,\,\left[ {M = \left( {{x_M};{y_M}} \right)} \right]\)


\(A,B\,\,\,\,\left\{ \matrix{
\,{\left( {x - 1} \right)^2} + {y^2} = {5^2}\,\,\,\,\,\left[ {{\rm{circle}}} \right] \hfill \cr
\,y = 2x\,\,\,\,\,\left[ {{\rm{line}}} \right] \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,{\left( {x - 1} \right)^2} + {\left( {2x} \right)^2} = {5^2}\,\,\,\, \Rightarrow \,\,\,\,5{x^2} - 2x - 24 = 0\,\,\,\,\left( * \right)\)


\(M = \left( {{{{x_A} + {x_B}} \over 2}\,\,;\,\,{{{y_A} + {y_B}} \over 2}} \right)\,\,\,\mathop = \limits^{M\, \in \,\,{\rm{line}}} \,\,\,\left( {{{{x_A} + {x_B}} \over 2}\,\,;\,\,{{2\,{x_A} + 2\,{x_B}} \over 2}} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,? = {3 \over 2}\left( {{x_A} + {x_B}} \right)\)


\(\left( * \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,{x_A} + {x_B} = {\rm{sum}}\,\,{\rm{roots}} = - {{\left( { - 2} \right)} \over 5} = {2 \over 5}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,? = {3 \over 2}\left( {{2 \over 5}} \right) = {3 \over 5}\)


The correct answer is therefore (C).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________

Assume that There is a circle and cord is formed with the points intercecting the circle with the line A and B.

Hence perpendicular from circle always intersects at the mid point. Just gind the perpendicular point from (1,0) on the line segment formed by origin and (4,8).


answer is 3/5

Posted from my mobile device

The circle with center (1,0) and radius 5 intercepts
eqnof rhe circle = (x-1)^2+y^2 = 25 ......1
eqn of the line straight line defined by C = (0,0) and D= (4,8)
y=2x......
solving from 1 and 2 provides
(2/5,24/5) and (-2,-4)
solving for midpoint we get
(1/5,2/5)
=> sum = 3/5 hence IMO C