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 $y=m x$ 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 $x^2+y^2-k y=0 \forall k \in R$
• 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 . $
$\text { Now }( OT )^2=( PT )^2 \text { (given) }$
$\Rightarrow( y - mx )^2= x ^2+ m ^2 x ^2$
$\Rightarrow y ^2+ m ^2 x ^2-2 mxy = x ^2+ m ^2 x ^2$
$\therefore \frac{ dy }{ dx }=\frac{ y ^2- x ^2}{2 xy }$
Put $y^2=t \Rightarrow 2 y \frac{d y}{d x}=\frac{d t}{d x} \Rightarrow x \frac{d t}{d x}=t-x^2 \Rightarrow \frac{d t}{d x}-\frac{t}{x}=-x$
$\therefore \text { I.F. }= e ^{\int-\frac{1}{x} dx }= e ^{-\ln x}=\frac{1}{ x }$
The solution is given by $t\left(\frac{1}{x}\right)=\int-d x=-x+C \Rightarrow t=-x^2+C x$
Hence $y ^2+ x ^2= Cx$.
If this is passing through $(2,2) \Rightarrow C=4 \Rightarrow x^2+y^2=4 x \Rightarrow(x-2)^2+y^2=4$
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.]