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 $a^2>k{b}^2$, where $k=$

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (A)

$2$

The given circle is $2x(x-a)+y(2y-b)=0\Rightarrow x^2+y^2-ax-\frac{by}{2}=0$

Center 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 $y-0=\frac{2(2h-a)}{b}(x-h)$

$\Rightarrow by=2(2h-a)(x-h)\Rightarrow (4h-2a)x-by-4h^2+2ah=0$

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

$(4h-a)a-b.\frac{b}{2}-4h^2+2ah=0\Rightarrow 8h^2-12ah+4a^2+b^2=0$

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

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

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