Circle Question 9
Question 9 - 31 January - Shift 2
The set of all values of $a^{4}$ for which the line $x+y=0$ bisects two distinct chords drawn from a point $P(\frac{1+a}{2}, \frac{1-a}{2})$ on the circle $2 x^{2}+2 y^{2}-(1+a) x-(1-a) y=0$ is equal to:
(1) $(8, \infty)$
(2) $(4, \infty)$
(3) $(0,4]$
(4) $(2,12]$
Show Answer
Answer: (1)
Solution:
Formula: Equation of chord
$x^{2}+y^{2}-\frac{(1+a) x}{2}-\frac{(1-a) y}{2}=0$
Centre $(\frac{1+a}{4}, \frac{1-a}{4}) \Rightarrow(h, k)$
$P(\frac{1+a}{2}, \frac{1-a}{2}) \Rightarrow(2 h, 2 k)$
Equation of chord $\Rightarrow T=S_1$
$\Rightarrow(x-y) \lambda-\frac{2 h(x+\lambda)}{2}-\frac{(2 k)(y-\lambda)}{2}$
$ =2 \lambda^{2}-2 h \lambda $
Now, $\lambda 2 h \quad$ satisfies the chord
$\therefore(2 h-2 k) \lambda-h(x+\lambda)-k(y-\lambda)$
$\Rightarrow 2 \lambda^{2}+4 k \lambda-4 h \lambda+h \lambda-k \lambda+hx+ky=0$
$\Rightarrow 2 \lambda^{2}+\lambda(3 k-3 h)+ky+hx=0$
$\Rightarrow D>0$
$\Rightarrow 9(k-h)^{2}-8(ky+hx)$
$\Rightarrow 7 k^{2}+7 h^{2}+18 kh<0$
$\Rightarrow 7(\frac{1-\mathbf{a}}{4})^{2}+7(\frac{1+a}{4})^{2}+18(\frac{1-a^{2}}{16})<0$
$\Rightarrow 7[\frac{2(1+a^{2})}{16}]+\frac{18(1-a^{2})}{16}<0, \quad a^{2}=t$
$\Rightarrow \frac{7}{8}(1+t)+\frac{18(1-t)}{16}<0$
$\Rightarrow \frac{14+14 t+18-18 t}{16}<0$
$\Rightarrow 4 t>32$
$t>8 \quad \text{since} \ t = a^2 $
$\therefore \quad a^{2}>8$
