The locus of the centre of the circle which cuts off an intercept of constant length on the  x-axis and which passes through a fixed point on  the y- axis is ?

  

 

Asked by Malavika Umesh | 20th May, 2015, 05:53: PM

Expert Answer:

Refer to the figure above. 
Let the fixed point on Y-axis be (0,p). The length of intercept on X-axis be 'L'.
Let the coordinates of center be (h,k)
From the figure, the radius of the circle is 
square root of k squared plus L squared over 4 end root
The equation of the circle can be written as
open parentheses x minus h close parentheses squared plus open parentheses y minus k close parentheses squared equals k squared plus L squared over 4
Since, the circle passes through the point (0,p), substituting the coordinates in the above equation, we get
h squared plus open parentheses p minus k close parentheses squared equals k squared plus L squared over 4 rightwards double arrow h squared plus p squared plus k squared minus 2 p k equals k squared plus L squared over 4 rightwards double arrow h squared plus p squared minus 2 p k equals L squared over 4 rightwards double arrow h squared equals 2 p k minus p squared plus L squared over 4
 
Replacing (h,k) by (x,y), we get
x squared equals 2 p y minus p squared plus L squared over 4
This is an equation linear in 'y' and quadratic in 'x'. Hence, the locus will be a parabola.

Answered by satyajit samal | 21st May, 2015, 10:33: AM