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  • Let a circle be given by 2x(x – a) + y(2y – b) = 0, a ≠ 0, b ≠ 0. Find the condition on a and b if two chords, each bisected by the x-axis can be drawn

Let a circle be given by 2x(x – a) + y(2y – b) = 0, a ≠ 0, b ≠ 0. Find the condition on a and b if two chords, each bisected by the x-axis can be drawn to the circle from \mathrm{\left ( a,\frac{b}{2} \right )}

 

Option: 1

a^{2}< 2b^{2}


Option: 2

a^{2}=2b^{2}


Option: 3

a^{2}> 2b^{2}


Option: 4

a^{2}> b^{2}


Answers (1)

best_answer

Given circle is 2x(x – a) + y(2y – b) = 0, a ≠ 0, b ≠ 0 orx^{2}+y^{2}-ax-\frac{b}{2}y=0

Let a=AA \equiv(a, b / 2). Clearly A is a point on the circle. Let a chord through A be bisected by the x-axis. Let this chord cuts the circle again at (α, β), then \mathrm{}\\frac{\beta+\frac{b}{2}}{2}=0 \Rightarrow \beta=-\frac{b}{2}

 Since, (α, β) lies on the circle,

\mathrm{\begin{aligned} & \alpha^2+\beta^2-a \alpha-\frac{b}{2} \beta=0 \\ & \Rightarrow a^2+\frac{b^2}{4}-a a+\frac{b^2}{4}=0 \\ & \Rightarrow 2 a^2-2 a \alpha+b^2=0 \\ & \end{aligned}}

for two such chords there would be two distinct real values of α.

So, D > 0

\mathrm{\Rightarrow 4 a^2-8 b^2>0 \text { or } a^2>2 b^2}

 

 

Posted by

Pankaj Sanodiya

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