Circle 5 Question 20
20. Let a circle be given by
$2 x(x-a)+y(2 y-b)=0,(a \neq 0, b \neq 0)$
Find the condition on $a$ and $b$ if two chords, each bisected by the $X$-axis can be drawn to the circle from $(a, b / 2)$.
$(1992,6 M)$
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Answer:
Correct Answer: 20. $a^{2}>2 b^{2}$
Solution:
- The given circle can be rewritten as
$ x^{2}+y^{2}-a x-\frac{b y}{2}=0 $
Let one of the chord through $(a, b / 2)$ be bisected at $(h, 0)$.
Then, the equation of the chord having $(h, 0)$ as mid-point is
$ \begin{gathered} T=S _1 \\ \Rightarrow h \cdot x+0 \cdot y-\frac{a}{2}(x+h)-\frac{b}{4}(y+0)=h^{2}+0-a h-0 \\ \Rightarrow \quad h-\frac{a}{2} \quad x-\frac{b y}{4}-\frac{a}{2} h=h^{2}-a h \end{gathered} $
It passes through $(a, b / 2)$, then
$ \begin{aligned} & h-\frac{a}{2} \quad a-\frac{b}{4} \cdot \frac{b}{2}-\frac{a}{2} h=h^{2}-a h \\ \Rightarrow \quad & h^{2}-\frac{3}{2} a h+\frac{a^{2}}{2}+\frac{b^{2}}{8}=0 \end{aligned} $
According to the given condition, Eq. (iii) must have two distinct real roots. This is possible, if the discriminant of Eq. (iii) is greater than 0 .
i.e. $\quad \frac{9}{4} a^{2}-4 \frac{a^{2}}{2}+\frac{b^{2}}{8}>0 \Rightarrow \frac{a^{2}}{4}-\frac{b^{2}}{2}>0$
$\Rightarrow \quad a^{2}>2 b^{2}$
