Question

Let a circle be given by $2x(x-a)+y(2y-b)=0(a\ne 0,b\ne 0)$. If two chords, each bisected by the x-axis, can be drawn to the circle from $(a,\frac{b}{2})$, then

• A. $a^2>2b^2$
• B. $a^2>b^2$
• C. $a^2<2b^2$
• D. $a^2<b^2$

Answer 1

The correct option is A $a^2>2b^2$

The given circle is $2x(x-a)+y(2y-b)=0$

or, $x^2+y^2-ax-\frac{by}{2}=0.$

Centre of this circle is $(\frac{a}{2},\frac{b}{4}).$

Let $AB$ be the chord which is bisected by x-axis at a point $M.$

Let its coordinates be $M(h,0).$

Slope of $CM=\frac{0-\frac{b}{4}}{h-\frac{a}{2}}=\frac{-b}{2(2h-a)}.$

Since $CM\perp AB,$

$\therefore$ Slope of $AB=\frac{2(2h-a)}{b}$

$\therefore$Equation of $AB$ is $y-0=\frac{2(2h-a)}{b}(x-h)$

or, $by=2(2h-a)(x-h)$

or, $(4h-2a)x-by-4h^2+2ah=0.$

Since it passes through $(a,\frac{b}{2}),$ we have

$(4h-2a)a-b.\frac{b}{2}-4h^2+2ah=0.$

or, $8h^2-12ah+4a^2+b^2=0.$

Now there are two chords disected by the x-axis, so there must be two distinct real roots of $h.$

$\therefore$ Discriminant $>0$ or ${(-12a)}^2-4.8(4a^2+b^2)>0$

$\Rightarrow 144a^2-32(4a^2+b^2)>0\Rightarrow a^2>2b^2.$