Circle 1 Question 18
18. Consider a curve $a x^{2}+2 h x y+b y^{2}=1$ and a point $P$ not on the curve. A line drawn from the point $P$ intersect the curve at points $Q$ and $R$. If the product $P Q \cdot Q R$ is independent of the slope of the line, then show that the curve is a circle.
$(1997,5 M)$
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Answer:
Correct Answer: 18. (d)
Solution:
- The given circle is $a x^{2}+2 h x y+b y^{2}=1$
Let the point $P$ not lying on Eq. (i) be $\left(x _1, y _1\right)$, let $\theta$ be the inclination of line through $P$ which intersects the given curve at $Q$ and $R$.
Then, equation of line through $P$ is
$$ \begin{aligned} \frac{x-x _1}{\cos \theta} & =\frac{y-y _1}{\sin \theta}=r \\ \Rightarrow \quad x=x _1+r \cos \theta, y & =y _1+r \sin \theta \end{aligned} $$
For points $Q$ and $R$, above point must lie on Eq. (i).
$\Rightarrow \quad a\left(x _1+r \cos \theta\right)^{2}+2 h\left(x _1+r \cos \theta\right)\left(y _1+r \sin \theta\right)$
$$ +b\left(y _1+r \sin \theta\right)^{2}=1 $$
$\Rightarrow\left(a \cos ^{2} \theta+2 h \sin \theta \cos \theta+b \sin ^{2} \theta\right) r^{2}$ $+2\left(a x _1 \cos \theta+h x _1 \sin \theta+h y _1 \cos \theta+b y _1 \sin \theta\right) r$
$$ +\left(a x _1^{2}+2 h x _1 y _1+b y _1^{2}-1\right)=0 $$
It is quadratic in $r$, giving two values of $r$ as $P Q$ and $P R$.
$\therefore \quad P Q \cdot P R=\frac{a x _1^{2}+2 h x _1 y _1+b y _1^{2}-1}{a \cos ^{2} \theta+2 h \sin \theta \cos \theta+b \sin ^{2} \theta}$ Here, $a x _1^{2}+2 h x _1 y _1+b y _1^{2}-1 \neq 0$, as $\left(x _1, y _1\right)$ does not lie on Eq. (i),
Also, $a \cos ^{2} \theta+2 h \sin \theta \cos \theta+b \sin ^{2} \theta$
$$ \begin{aligned} & =a+2 h \sin \theta \cos \theta+(b-a) \sin ^{2} \theta \\ & =a+\sin \theta{2 h \cos \theta+(b-a) \sin \theta} \\ & =a+\sin \theta \cdot \sqrt{4 h^{2}+(b-a)^{2}} \cdot(\cos \theta \sin \varphi+\sin \theta \cos \varphi) \end{aligned} $$
where, $\tan \theta=\frac{b-a}{2 h}$
$$ =a+\sqrt{4 h^{2}+(b-a)^{2}} \sin \theta \sin (\theta+\varphi) $$
which will be independent of $\theta$, if
$$ \begin{aligned} & 4 h^{2}+(b-a)^{2}=0 \\ & \Rightarrow \quad h=0 \quad \text { and } \quad b=a \end{aligned} $$
$\therefore$ Eq. (i) reduces to $x^{2}+y^{2}=\frac{1}{a}$
which is a equation of circle.
