Consider the family of all circles whose cent | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The equation of a circle with center $(h, h)$ and radius $r$ is $(x - h)^2 + (y - h)^2 = r^2$.Expanding this,we get $x^2 - 2xh + h^2 + y^2 - 2yh +

Vedclass · Accepted Answer

$(B, C)$

Vedclass · Answer

(C) The equation of a circle with center $(h, h)$ and radius $r$ is $(x - h)^2 + (y - h)^2 = r^2$.Expanding this,we get $x^2 - 2xh + h^2 + y^2 - 2yh + h^2 = r^2$,which simplifies to $x^2 + y^2 - 2h(x + y) + 2h^2 - r^2 = 0$.Let $C = 2h^2 - r^2$. Then $x^2 + y^2 - 2h(x + y) + C = 0$.Differentiating with respect to $x$: $2x + 2yy^{\prime} - 2h(1 + y^{\prime}) = 0$,which gives $h = \frac{x + yy^{\prime}}{1 + y^{\prime}}$.Substituting $h$ back into the equation: $x^2 + y^2 - 2\left(\frac{x + yy^{\prime}}{1 + y^{\prime}}\right)(x + y) + C = 0$.Differentiating again: $2 + 2(y^{\prime})^2 + 2yy^{\prime \prime} - 2\frac{d}{dx}\left[\frac{(x + yy^{\prime})(x + y)}{1 + y^{\prime}}\right] = 0$.After simplification,the differential equation is $(y - x)y^{\prime \prime} + (1 + y^{\prime} + (y^{\prime})^2)y^{\prime} + 1 = 0$.Comparing this with $Py^{\prime \prime} + Qy^{\prime} + 1 = 0$,we get $P = y - x$ and $Q = 1 + y^{\prime} + (y^{\prime})^2$.Thus,$P = y - x$ (Statement $B$ is true).$P - Q = (y - x) - (1 + y^{\prime} + (y^{\prime})^2) = y - x - 1 - y^{\prime} - (y^{\prime})^2$ (Statement $D$ is false).$P + Q = (y - x) + (1 + y^{\prime} + (y^{\prime})^2) = 1 - x + y + y^{\prime} + (y^{\prime})^2$ (Statement $C$ is true).Therefore,the correct option is $(B, C)$.

Similar Questions

$A$ solution curve of the differential equation $(x^2+xy+4x+2y+4) \frac{dy}{dx}-y^2=0, x>0$,passes through the point $(1,3)$. Then the solution curve

Verify that the given function $y = x \sin x$ is a solution of the differential equation $x y^{\prime} = y + x \sqrt{x^2 - y^2}$ (where $x \neq 0$ and $x > y$ or $x < -y$).

The equation of the curve passing through $(3, 4)$ and satisfying the differential equation $y \left( \frac{dy}{dx} \right)^2 + (x - y) \frac{dy}{dx} - x = 0$ can be:

The solution of $(xy \cos xy + \sin xy)dx + x^2 \cos xy \, dy = 0$ is

Find the family of curves such that the angle between the tangent at any point $(x, y)$ and the tangent to the curve $xy = c^2$ at the point of intersection is $\frac{\pi}{4}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module