SATHEE: Circle 5 Question 19

Circle 5 Question 19

19. Find the intervals of values of $a$ for which the line $y + x = 0$ bisects two chords drawn from a point $\frac{1 + \sqrt{2} a}{2}, \frac{1 - \sqrt{2} a}{2}$ to the circle

$2 x^{2} + 2 y^{2} - (1 + \sqrt{2} a) x - (1 - \sqrt{2} a) y = 0$ .

$(1996, 6 M)$

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Answer:

Correct Answer: 19. $a \in (- \infty, - 2) \cup (2, \infty)$

Solution:

Given, $2 x^{2} + 2 y^{2} - (1 + \sqrt{2} a) x - (1 - \sqrt{2} a) y = 0$

$\Rightarrow x^{2} + y^{2} - \frac{1 + \sqrt{2} a}{2} x - \frac{1 - \sqrt{2} a}{2} y = 0$

Since, $y + x = 0$ bisects two chords of this circle, mid points of the chords must be of the form $(α, - α)$ .

Equation of the chord having $(α, - α)$ as mid points is

$T = S_{1}$

$\Rightarrow x α + y (- α) - \frac{1 + \sqrt{2} a}{4} (x + α) - \frac{1 - \sqrt{2} a}{4} (y - α)$

$= α^{2} + (- α)^{2} - \frac{1 + \sqrt{2} a}{2} α - \frac{1 - \sqrt{2} a}{2} (- α)$

$\Rightarrow 4 x α - 4 y α - (1 + \sqrt{2} a) x - (1 + \sqrt{2} a) α$

$- (1 - \sqrt{2} a) y + (1 - \sqrt{2} a) α$

$= 4 α^{2} + 4 α^{2} - (1 + \sqrt{2} a) \cdot 2 α + (1 - \sqrt{2} a) \cdot 2 α$

$\Rightarrow 4 α x - 4 α y - (1 + \sqrt{2} a) x - (1 - \sqrt{2} α) y$

$= 8 α^{2} - (1 + \sqrt{2} a) α + (1 - \sqrt{2} a) α$

But this chord will pass through the point

$\frac{1 + \sqrt{2} a}{2}, \frac{1 - \sqrt{2} a}{2}$

$4 α \frac{1 + \sqrt{2} a}{2} - 4 α \frac{1 - \sqrt{2} a}{2} - \frac{(1 + \sqrt{2} a) (1 + \sqrt{2} a)}{2}$

$- \frac{(1 - \sqrt{2} a) (1 - \sqrt{2} a)}{2}$

$= 8 α^{2} - 2 \sqrt{2} α α$

$\Rightarrow 2 α [(1 + \sqrt{2} a - 1 + \sqrt{2} α)] = 8 α^{2} - 2 \sqrt{2} a α$

$\Rightarrow 4 \sqrt{2} α α - \frac{1}{2} [2 + 2 (\sqrt{2} a)^{2}] = 8 α^{2} - 2 \sqrt{2} α α$

$[∵ (a + b)^{2} + (a - b)^{2} = 2 a^{2} + 2 b^{2}]$

$\Rightarrow 8 α^{2} - 6 \sqrt{2} a α + 1 + 2 a^{2} = 0$

But this quadratic equation will have two distinct roots, if

$\begin{array}{ll} (6 \sqrt{2} a)^{2} - 4 (8) (1 + 2 a^{2}) > 0 & \Rightarrow 72 a^{2} - 32 (1 + 2 a^{2}) > 0 \\ \Rightarrow & 8 a^{2} - 32 > 0 \\ \Rightarrow & \Rightarrow< - 2 \cup a^{2} - 4 > 0 \end{array}$

Therefore, $a \in (- \infty, - 2) \cup (2, \infty)$ .

Circle 5 Question 19

19. Find the intervals of values of $a$ for which the line $y + x = 0$ bisects two chords drawn from a point $\frac{1 + \sqrt{2} a}{2}, \frac{1 - \sqrt{2} a}{2}$ to the circle

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Circle 5 Question 19

19. Find the intervals of values of a for which the line y+x=0 bisects two chords drawn from a point 1+2a2,1−2a2 to the circle

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19. Find the intervals of values of $a$ for which the line $y + x = 0$ bisects two chords drawn from a point $\frac{1 + \sqrt{2} a}{2}, \frac{1 - \sqrt{2} a}{2}$ to the circle