Circle Ques 93
- If two distinct chords, drawn from the point $(p, q)$ on the circle $x^{2}+y^{2}=p x+q y$ (where, $p q \neq 0$ ) are bisected by the $X$-axis, then
(a) $p^{2}=q^{2}$
(c) $p^{2}<8 q^{2}$
(b) $p^{2}=8 q^{2}$
(d) $p^{2}>8 q^{2}$
$(1999,2 M)$
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Answer:
Correct Answer: 93.(d)
Solution:
Formula:
Important Points to be Remembered
- 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.
$A \equiv(p, q) \cdot C$ is mid-point and coordinate of $C$ is $(h, 0)$ Then, coordinates of $B$ are $(-p+2 h,-q)$ and $B$ lies on the circle $x^{2}+y^{2}=p x+q y$, we have
$$ \begin{array}{rlrl} & & (-p+2 h)^{2}+(-q)^{2} & =p(-p+2 h)+q(-q) \\ \Rightarrow & p^{2}+4 h^{2}-4 p h+q^{2} & =-p^{2}+2 p h-q^{2} \\ \Rightarrow & 2 p^{2}+2 q^{2}-6 p h+4 h^{2} & =0 \\ \Rightarrow & 2 h^{2}-3 p h+p^{2}+q^{2} & =0 \end{array} $$
There are given two distinct chords which are bisected at $X$-axis then, there will be two distinct values of $h$ satisfying Eq. (i).
So, discriminant of this quadratic equation must be $>0$.
$$ \begin{array}{lc} \Rightarrow & D>0 \\ \Rightarrow & (-3 p)^{2}-4 \cdot 2\left(p^{2}+q^{2}\right)>0 \\ \Rightarrow & 9 p^{2}-8 p^{2}-8 q^{2}>0 \\ \Rightarrow & p^{2}-8 q^{2}>0 \Rightarrow p^{2}>8 q^{2} \end{array} $$
BETA
