JEE Main 2022 — Differential Equations Question with Solution

$$\frac{dy}{dx}=\frac{ax-by+a}{bx+cy+a}$$

$$=bxdy+cydy+ady=axdx-bydx+adx$$

$$=cydy+ady-axdx-adx+b(xdy+ydx)=0$$

$$=c\int ydy+a\int xdx-a\int dx+b\int d(xy)=0$$

$$=\frac{c{y}^2}{2}+ay-\frac{a{x}^2}{2}-ax+bxy=k$$

$$=a{x}^2-c{y}^2+2ax-2ay-2bxy=k$$

Above equation is circle

$$\Rightarrow$$ a = $$-$$ c and b = 0

$$a{x}^2+a{y}^2+2ax-2ay=k$$

$$\Rightarrow x^2+y^2+2x-2y=\lambda \left[\lambda =\frac{k}{a}\right]$$

Passes through (2, 5)

$$4+25+4-10=\lambda \Rightarrow \lambda =23$$

Circle $$\equiv x^2+y^2+2x-2y-23=0$$

Centre ($$-$$1, 1) $$r=\sqrt{{(-1)}^2+1^2+23}=5$$

Shortest distance of $$(11,6)=\sqrt{{12}^2+5^2}-5$$

$$=13-5$$

$$=8$$