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The differential equation of the family of circles with fixed radius $$5$$ units and centre on the line $$y = 2$$ is :

JEE · Math · previous-year question

  1. A.$$\left( {x - 2} \right){y^2} = 25 - {\left( {y - 2} \right)^2}$$
  2. B.$$\left( {y - 2} \right){y^2} = 25 - {\left( {y - 2} \right)^2}$$
  3. C.$${\left( {y - 2} \right)^2}{y^2} = 25 - {\left( {y - 2} \right)^2}$$correct
  4. D.$${\left( {x - 2} \right)^2}{y^2} = 25 - {\left( {y - 2} \right)^2}$$

Answer

C. $${\left( {y - 2} \right)^2}{y^2} = 25 - {\left( {y - 2} \right)^2}$$

Explanation

Let the center of the circle be $$(h, 2)$$ $$\therefore$$ Equation of circle is $${\left( {x - h} \right)^2} + \left( {y - 2} \right){}^2 = 25\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$ Differentiating with respect to $$x,$$ we get $$2\left( {x - h} \right) + 2\left( {y - 2} \right){{dy} \over {dx}} = 0$$ $$ \Rightarrow x - h = - \left( {y - 2} \right){{dy} \over {dx}}$$ Substituting in equation $$(1)$$ we get $${\left( {y - 2} \right)^2}{\left( {{{dy} \over {dx}}} \right)^2} + {\left( {y - 2} \right)^2} = 25$$ $$ \Rightarrow {\left( {y - 2} \right)^2}{\left( {y'} \right)^2} = 25 - {\left( {y - 2} \right)^2}$$

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