Differential Equations — Class 12 Maths Solution

ncert exercise SA NCERT Ex.9.3,Q.6,Page 391
Question

Form the differential equation of the family of circles touching the y-axis at origin

Step-by-step Solution

.: Let $C$ denote the family of circles touching
y-axis at the origin.

Let $\left( {a,0} \right)$
be the co-ordinates of the centre of any member
of the family.

Therefore, equation of family $C$ is
${\left( {x - a} \right)^2} + {y^2} = {a^2}$
or ${x^2} + {y^2} = 2ax$

…(1)
where, a is any arbitrary constant.
Differentiating (1) w.r.t. $x$,

we get
$2x + 2y\cfrac{{dy}}{{d\mathfrak{r}}} = 2a \Rightarrow x + y\cfrac{{dy}}{{dx}} = a$

…(2)
Substituting the value of $a$ from (2) in (1),

we get
${x^2} + {y^2} = 2x\left( {x + y\cfrac{{dy}}{{dx}}} \right) \Rightarrow {x^2} + {y^2} = 2{x^2} + 2xy\cfrac{{dy}}{{dx}}$
$\Rightarrow {x^2} + {y^2} - 2{x^2} = 2xy{y_1} \Rightarrow 2xy{y_1} + {x^2} = {y^2}$

which is required differential equaiton.

NCERT & Exemplar solution for CBSE Class 12 Mathematics, Differential Equations. Curated by Sachin Sharma. Free for all students.