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Vector Algebra

Question
CBSEENMA12032965
Wired Faculty App

Form the differential equation of the family of circles having centre on x-axis and passing through the origin.    

Solution

Let the radius of circle = a
Since the circle has centre on x-axis and passes through the origin.
∴ its centre is (a, 0)
The equation of circle is
( x – a)2+( y – 0)2 = a2  or x2 –2 a x + a2 + y2 = a2
∴       x2 + y2 – 2 ax = 0    ...(1)
Differentiating both sides w.r.t. x, we get.
                           2 straight x plus 2 straight y dy over dx minus 2 straight a space equals space 0 space space space space space space space space space space space space or space space space space space space straight x plus straight y dy over dx minus straight a space equals space 0
therefore space space space space space space space space space space space straight a space equals space straight x plus straight y dy over dx
Putting this value of a in (1), we get.
                             straight x squared plus straight y squared minus 2 straight x open parentheses straight x plus straight y dy over dx close parentheses space equals space 0
therefore space space space straight x squared plus straight y squared minus 2 straight x squared minus 2 xy dy over dx space equals space 0 therefore space space space space minus straight x squared plus straight y squared minus 2 xy dy over dx space equals space 0 therefore space space space space space space 2 space xy dy over dx plus straight x squared minus straight y squared space equals space 0
which is required differential equation. 

Some More Questions From Vector Algebra Chapter

Determine order and degree (if defined) of differential equations:
y' + y = ex

Determine order and degree (if defined) of differential equations:
y'' + (y')2 + 2 y = 0

Determine order and degree (if defined) of differential equations:
y″ + 2y′ + sin y = 0

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