Question

If two distinct chords, drawn from the point (p, q) on the circle $x^2 + y^2 = px + qy$ (where, $ pq \ne 0$) are bisected by the X-axis, then

• A. $p^2 = q^2$
• B. $p^2 = 8q^2$
• C. $p^2 < 8q^2$
• D. $p^2 > 8q^2$

Answer 1

The Correct Option is B

NOTE In solving a line and a circle there oftengenerate a
quadratic equation and further we have to apply
condition of Discriminant so question convert from
coordinate to quadratic equation
From equation of circle it is clear that circle passes
through origin. Let A B is chord of the circle.
From equation of circle it is clear that circle passes
through origin. Let A B is chord of the circle.
A = (p, q). C is mid-point and coordinate of C is (h, 0)
Then, coordinates of B are (- p+ 2h, - q) and B lies on
the circle $x^2 + y^2 = px + qy$, we have
$(- P + 2h^2 + (- q)^2 = p\, ( - p + 2h) + q\, (- q)$
$\Rightarrow\hspace15mmp^2 + 4h^2 -4ph + q^2 = -p^2 +2ph - q^2 $
$\Rightarrow\hspace12mm2p^2 + 2q^2 - 6ph + 4h^2=0$
$\Rightarrow\hspace14mm2p^2 + 3ph + p^2 + q^2\, =0\, \, \, \, \, \, \, \, \, ...(i)$
There are given two distinct chords which are bisected
at X-axis then, there will be two distinct values of h
satisfying E (i).
So, discrim inant of this quadratic equation m ust be > 0
$\Rightarrow\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, D>0$
$\Rightarrow\hspace2mm(-3p)^2 - 4.2 (p^2 + q^2) >0$
$\Rightarrow\, \, \, \, \, \, \, \, \, 9p^2 - 8p^2 - 8q^2 D>0$
$\Rightarrow\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, p^2 - 8q^2>0$
$\Rightarrow\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, p^2 - 8q^2$