Question

The tangent at any point $P$ on a curve $f(x,y)=0$ cuts the $y$-axis at $T$. If the distance of the point from $P$ equals the distance of $T$ from the origin then the curve with this property represents a family of circles. Which of the following is/are CORRECT?

• A. Any arbitrary line $y=mx$ cuts every member of this family at the points where the slopes of these members are equal.
• B. $f(x,y)=0$ is orthogonal to the family of circles $x^2+y^2-ky=0\forall k\in R$
• C. If $f(x,y)=0$ passes through $(2,2)$ then the intercept made by its director circle on the $y$-axis is equal to 8 .
• D. If $f(x,y)=0$ passes through $(-1,1)$ then image of its centre in the line $y=x$, is $(1,0)$

Answer 1

Answer: (A), (B)

We have equation of tangent is $(Y-y)=m(X-x)$

$\text{ Put }X=0\text{, we get }Y=y-mx.$
$\text{ Now }(OT{)}^2=(PT{)}^2\text{ (given) }$
$\Rightarrow (y-mx{)}^2=x^2+m^2{x}^2$
$\Rightarrow y^2+m^2{x}^2-2mxy=x^2+m^2{x}^2$
$\therefore \frac{dy}{dx}=\frac{y^2-x^2}{2xy}$
Put $y^2=t\Rightarrow 2y\frac{dy}{dx}=\frac{dt}{dx}\Rightarrow x\frac{dt}{dx}=t-x^2\Rightarrow \frac{dt}{dx}-\frac{t}{x}=-x$
$\therefore \text{ I.F. }=e^{\int -\frac{1}{x}dx}=e^{-\ln x}=\frac{1}{x}$
The solution is given by $t\left(\frac{1}{x}\right)=\int -dx=-x+C\Rightarrow t=-x^2+Cx$
Hence $y^2+x^2=Cx$.
If this is passing through $(2,2)\Rightarrow C=4\Rightarrow x^2+y^2=4x\Rightarrow (x-2{)}^2+y^2=4$
It's director circle will be $(x-2{)}^2+y^2=8$.
Put $x=0,y^2=4\Rightarrow y=2$ or -2
Intercept on $y$-axis is 4 .
This represents a family of circles with centre at $y$-axis and passing through origin. Verify other alternatives.]