Animated Solution for Mathematics - Circles: If the circles $x^2+y^2+2x+2ky+6=0,x^2+y^2+2ky+k=0$ intersect orthogonally, then $k$ is

A Beautiful Intersection

Imagine you are standing in a vast, empty space, and suddenly, two circles appear. They are not just touching; they are crossing each other.

If you were to draw a tangent line to each circle at the exact point where they meet, you would find that these two lines are perfectly perpendicular to each other, meeting at a crisp $90^{\circ }$ angle. This is what we call orthogonal circles.

In the world of JEE Advanced, this isn't just a pretty picture; it is a powerful condition that allows us to unlock hidden variables.

To solve this, we don't need to draw complex diagrams. We have a secret weapon: the condition for orthogonality.

For any two circles given by the equations $x^2+y^2+2g_{1}x+2f_{1}y+c_1=0$ and $x^2+y^2+2g_{2}x+2f_{2}y+c_2=0$, the condition that they intersect orthogonally is elegantly simple:

This formula is a direct consequence of the Pythagorean theorem applied to the triangle formed by the centers of the circles and their point of intersection. It is a cornerstone of coordinate geometry, and once you master it, you hold the key to many complex problems.

Let us look at our specific problem. We have two circles: $x^2+y^2+2x+2ky+6=0$ and $x^2+y^2+2ky+k=0$.

Our first step is to extract the parameters $g,f,$ and $c$ by comparing them to the standard form $x^2+y^2+2gx+2fy+c=0$.

For the first circle, $x^2+y^2+2x+2ky+6=0$, we immediately see that $2g_1=2$, so $g_1=1$. Similarly, $2f_1=2k$, so $f_1=k$, and the constant $c_1=6$.

For the second circle, $x^2+y^2+2ky+k=0$, we must be careful. There is no $x$ term, which means the coefficient of $x$ is $0$.

Thus, $2g_2=0$, which gives us $g_2=0$. The $y$ term is $2ky$, so $f_2=k$, and the constant $c_2=k$. This is where many students stumble—do not let the missing $x$ term trick you!

Now, we simply plug these values into our orthogonality condition:

The first term, $2(1)(0)$, vanishes into zero. We are left with the following quadratic equation:

To solve this, we split the middle term: $2k^2-4k+3k-6=0$. Factoring by grouping, we get:

This simplifies to $(2k+3)(k-2)=0$. Setting each factor to zero, we find our solutions:

You have just navigated the logic of orthogonal circles with precision. Remember, geometry is just algebra in disguise, and algebra is just geometry waiting to be solved.