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Joint Entrance Examination
Mathematics
Circle
Question 172
Question 172 of 219
Medium
MCQ
The differential equation of the family of circles with fixed radius
5
5
5
units and centre on the line
y
=
2
y = 2
y
=
2
is :
A
\[\left( {x - 2} \right){y^2} = 25 - {\left( {y - 2} \right)^2}\]
B
\[\left( {y - 2} \right){y^2} = 25 - {\left( {y - 2} \right)^2}\]
C
\[{\left( {y - 2} \right)^2}{y^2} = 25 - {\left( {y - 2} \right)^2}\]
D
\[{\left( {x - 2} \right)^2}{y^2} = 25 - {\left( {y - 2} \right)^2}\]
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Official solution
Let the center of the circle be
(
h
,
2
)
(h, 2)
(
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
,
x,
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
)
(1)
(
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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