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Please help me, I have no idea how to proceed.

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    I do not understand "if the $y-axis divides both (the) chords in the ratio 2:3". – 2017-01-02
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    Well the y axis divides the chords in the ratio 2:3 then which of the conditions in the options are met? – 2017-01-02
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    @JeanMarie I take it to mean that the $x$-coordinates of each chord are of the form $2\alpha$ and $-3\alpha$, and similarly for the difference in $y$-coordinates. – 2017-01-02
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    The problem statement is a bit ambiguous, I think. If the short side of the chords is on the same side as $(h,k)$ you get a different constraint on $h$ and $k$ than if the short side is opposite $(h,k)$. – 2017-01-02

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Assuming that the ratio as stated means that the shorter side of the chords is on the same side of the $y$-axis as the point $(h,k)$, the $x$-coordinate of the other end of the two chords must have an $x$-coordinate of $-\frac32h$. Solving for the corresponding $y$ coordinates will give you a condition on $h$ and $k$ for the resulting quadratic to have two solutions.

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    How can we be sure that that shorter side is on the same side of y- axis as the point (h,k)? – 2017-01-03
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    @Siddharth If you instead interpret the ratio as meaning that the long side is with $P$ instead, none of the presented solutions match the resulting constraint on $h$ and $k$. That, and the fact that the problem describes the chords as being drawn *from* $P$ supports the short side interpretation. – 2017-01-03
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    @Siddharth BTW, the solution to this problem comes down to determining what values of $h$ and $k$ result in two intersections between the line $x=-\frac32h$ and the circle. Since the segment from $P$ to the origin is a diameter of the circle, this suggests simply looking at $x$ coordinates. If $h\gt0$, then we must have $\frac12(h-\sqrt{h^2+k^2})\lt-\frac32h$ for the vertical line to intersect the circle at two points, and similarly for $h\lt0$. Solving for $k$ from there is simple. – 2017-01-03
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We can assume WLOG, that, up to symmetries with respect to coordinate axes, $h > 0$ and $k \geq 0$ (the case $k=0$ has no interest).

The intersection points of the circle and the straight line with slope $m$ verify the system:

$$\begin{cases}x^2+y^2=k x+ h y\\y-l=m(x-k)\end{cases}$$

Thus, the abscissas of these points verify the quadratic equation

$$x^2+(mx + l -mk)^2-kx-h(mx + l -mk)=0$$

whose solutions in $x$ are

  • $x_1=h$ (evident solution because $P(h,k)$ is a point of the circle) and

  • $x_2=\dfrac{hm^2-mk}{1+m^2}.$

Because $x_1 > 0$ (see first sentence), we have to assume

$$\tag{1}x_2 <0 \ \ \iff \ \ m<\dfrac{k}{h}.$$

Let us now express the ratio condition 3:2. By thaled property, it suffices to express it on the abscissas, taking care that it must be done with the absolute value of $|x_2|$, i.e., its opposite:

$$\dfrac{(mk-hm^2)/(1+m^2)}{h}=\dfrac{3}{2} \ \ \iff$$

$$2(mk-hm^2)=3h(1+m^2) \ \ \iff$$

$$\tag{2} (5h)m^2+(-2k)m+3h=0$$

The condition to be fulfilled is that there must exist two straight lines with this property, i.e., two slopes. Otherwise said, the quadratic equation ($1$) in variable $m$ must have 2 real solutions.

An equivalent condition is that its discriminant $\Delta=4k^2-60h^2$ is $>0$

We find thus $k^2>15h^2$, i.e. condition (A).

This can be seen as a necessary condition. Nevertheless, there is a pending condition $(1)$ that must also be fulfilled.

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    The problem statement forces $x_2\lt0$: the chords are divided by the $y$-axis. – 2017-01-03
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    I completely agree that the this condition is necessary for the reason you say. But I think that proceeding like this by a necessary conditon is ... not sufficient (!). We have answered a question that could have been : find a condition $k^2>15h^2$ under which a solution might exist But now, does this solution exists ? Unfortunately, I have not time enough these days to deepen the question... – 2017-01-03
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    The problem statement tacitly posits the existence of a solution: “Two chords **are drawn**...” – 2017-01-03