Two chords are drawn from the point (h,k) on the circle $x^2 +y^2=hx+ky$ . The y axis divides both the chords in the ratio 2:3

Question

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

Answer 1

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.

Answer 2

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.