The set of values of $a$ for which the point $(a - 1, a + 1)$ lies outside the circle $x^{2} + y^{2} = 8$ and inside the circle $x^{2} + y^{2} - 12 x + 12 y - 62 = 0$ is

(- 3 \sqrt{2}, - \sqrt{3}) \cup (\sqrt{3}, 3 \sqrt{2})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(- 2 \sqrt{2}, - \sqrt{3}) \cup (\sqrt{3}, 3 \sqrt{2})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(- 3 \sqrt{2}, \sqrt{3}) \cup (\sqrt{3}, 3 \sqrt{2})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(- 3 \sqrt{2}, - \sqrt{3}) \cup (- \sqrt{3}, 3 \sqrt{2})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $(- 3 \sqrt{2}, - \sqrt{3}) \cup (\sqrt{3}, 3 \sqrt{2})$
Point $(a - 1, a + 1)$ lies outside the circle $x^{2} + y^{2} = 8$
$\Rightarrow (a - 1)^{2} + (a + 1)^{2} - 8 > 0 \Rightarrow a^{2} - 3 > 0 a \in (- \infty, - \sqrt{3}) \cup (\sqrt{3} . \infty) \dots (1)$

Point $(a - 1, a + 1)$ lies inside the circle $(x - 6)^{2} + (y + 6)^{2} - 134 = 0$
$\Rightarrow (a - 7)^{2} + (a + 7)^{2} - 134 < 0$
$\Rightarrow a^{2} - 18 < 0$
$\Rightarrow a \in (- 3 \sqrt{2}, 3 \sqrt{2}) \dots (2)$

From $(1)$ and $(2)$

$a \in (- 3 \sqrt{2}, - \sqrt{3}) \cup (\sqrt{3}, 3 \sqrt{2})$

Suggest Corrections

Similar questions

x^{2} + y^{2} = 1

a

(a - 1, a + 1)

x^{2} + y^{2} = 8

x^{2} + y^{2} - 12 x + 12 y - 62 = 0

A = {1, 2, 3}

R_{1} = {(1, 1), (2, 2)}

R_{2} = {(1, 1), (2, 2) (3, 3), (1, 3)}

R_{3} = {(1, 1), (2, 2), (3, 3)}

x^{2} - 3 x + 3

(- 3, \frac{3}{2})

Join BYJU'S Learning Program