SATHEE: Circle 1 Question 18

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 $(x_{1}, y_{1})$ , let $θ$ be the inclination of line through $P$ which intersects the given curve at $Q$ and $R$ .

Then, equation of line through $P$ is

$\begin{aligned} \frac{x - x_{1}}{\cos θ} & = \frac{y - y_{1}}{\sin θ} = r \\ \Rightarrow x = x_{1} + r \cos θ, y & = y_{1} + r \sin θ \end{aligned}$

For points $Q$ and $R$ , above point must lie on Eq. (i).

$\Rightarrow a {(x_{1} + r \cos θ)}^{2} + 2 h (x_{1} + r \cos θ) (y_{1} + r \sin θ)$

$+ b {(y_{1} + r \sin θ)}^{2} = 1$

$\Rightarrow (a \cos^{2} θ + 2 h \sin θ \cos θ + b \sin^{2} θ) r^{2}$ $+ 2 (a x_{1} \cos θ + h x_{1} \sin θ + h y_{1} \cos θ + b y_{1} \sin θ) r$

$+ (a x_{1}^{2} + 2 h x_{1} y_{1} + b y_{1}^{2} - 1) = 0$

It is quadratic in $r$ , giving two values of $r$ as $P Q$ and $P R$ .

$∴ P Q \cdot P R = \frac{a x_{1}^{2} + 2 h x_{1} y_{1} + b y_{1}^{2} - 1}{a \cos^{2} θ + 2 h \sin θ \cos θ + b \sin^{2} θ}$ Here, $a x_{1}^{2} + 2 h x_{1} y_{1} + b y_{1}^{2} - 1 \neq 0$ , as $(x_{1}, y_{1})$ does not lie on Eq. (i),

Also, $a \cos^{2} θ + 2 h \sin θ \cos θ + b \sin^{2} θ$

$\begin{aligned} = a + 2 h \sin θ \cos θ + (b - a) \sin^{2} θ \\ = a + \sin θ 2 h \cos θ + (b - a) \sin θ \\ = a + \sin θ \cdot \sqrt{4 h^{2} + (b - a)^{2}} \cdot (\cos θ \sin φ + \sin θ \cos φ) \end{aligned}$