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Question

(A) The center of the circle touching the $y$-axis at the origin lies on the $x$-axis. Let $(a, 0)$ be the center of the circle.Since it touches the $

Vedclass · Accepted Answer

$x^2 - y^2 + 2xyy' = 0$

Vedclass · Answer

(A) The center of the circle touching the $y$-axis at the origin lies on the $x$-axis. Let $(a, 0)$ be the center of the circle.Since it touches the $y$-axis at the origin,its radius is $|a|$.The equation of the circle with center $(a, 0)$ and radius $|a|$ is $(x-a)^2 + y^2 = a^2$.Expanding this,we get $x^2 - 2ax + a^2 + y^2 = a^2$,which simplifies to $x^2 + y^2 = 2ax$ ... $(1)$.Differentiating equation $(1)$ with respect to $x$,we get $2x + 2yy' = 2a$,or $x + yy' = a$.Substituting the value of $a$ into equation $(1)$,we get $x^2 + y^2 = 2(x + yy')x$.$x^2 + y^2 = 2x^2 + 2xyy'$.Rearranging the terms,we get $y^2 - x^2 + 2xyy' = 0$,or $x^2 - y^2 = 2xyy'$.Thus,the required differential equation is $x^2 - y^2 + 2xyy' = 0$.

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

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