Test Index

Circles Part 2

Section: Mathematics

© examsnet.com

Question : 56 of 87

Marks: +1, -0

The set of all values of

a^2

for which the line

x+y=0

bisects two distinct chords drawn from

a

P \left(\frac{1+a}{2}, \frac{1-a}{2}\right)

2x^2+2y^2-(1+a)x-(1-a)y=0

[31-Jan-2023 Shift 2]

$(8,\infty)$
$(4,\infty)$
$(0,4]$
$(2,12]$

Solution:

\;x^2+y^2-\frac{(1+a)x}{2}-\frac{(1-a)y}{2}=0

\;\;\text{ Centre }\;\left(\frac{1+a}{4}, \frac{1-a}{4}\right) \Rightarrow (h,k)

\; P \left(\frac{1+a}{2}, \frac{1-a}{2}\right) \Rightarrow (2h, 2k)

Equation of chord

\Rightarrow T = S_1

\; \Rightarrow (x-y)\lambda - \frac{2h(x+\lambda)}{2} - \frac{(2k)(y-\lambda)}{2}

\; = 2\lambda^2 - 2h(\lambda) + 2k\lambda

Now,

\lambda (2h, 2k)

satisfies the chord

\; \therefore (2h-2k)\lambda - h(x+\lambda) - k(y-\lambda)

\; \Rightarrow 2\lambda^2 + 4k\lambda - 4h\lambda + h\lambda - k\lambda + hx + ky = 0

\; \Rightarrow 2\lambda^2 + \lambda(3k-3h) + ky + hx = 0

\; \Rightarrow D > 0

\; \Rightarrow 9(k-h)^2 - 8(ky + hx) > 0

\; \Rightarrow 9(k-h)^2 - 8(2k^2 + 2h^2) > 0

\; \Rightarrow -7k^2 - 7h^2 - 18kh > 0

\; \Rightarrow 7k^2 + 7h^2 + 18kh < 0

\; \Rightarrow 7\left(\frac{1-a}{4}\right)^2 + 7\left(\frac{1+a}{4}\right)^2 + 18\left(\frac{1-a^2}{16}\right) < 0

\; \Rightarrow 7\left[\frac{2(1+a^2)}{16}\right] + \frac{18(1-a^2)}{16} < 0, \;\; a^2 = t

\; \Rightarrow \frac{7}{8}(1+t) + \frac{18(1-t)}{16} < 0

\; \Rightarrow \frac{14+14t+18-18t}{16} < 0

\; \Rightarrow 4t > 32

\; t > 8 \;\; a^2 > 8

© examsnet.com

Go to Question:

Prev Question

Next Question