Animated Solution for Mathematics - Circles: The circles $x^2+y^2-10x+16=0$ and $x^2+y^2=r^2$ intersect each other in two distinct points if

A Geometric Journey

Welcome, my dear students! Today, we are going to explore a problem that is a classic in the world of JEE geometry. It is not just about crunching numbers; it is about visualizing the beautiful, rhythmic dance of two circles in a coordinate plane.

Imagine you are standing on a vast, flat plane. You have one circle, fixed and unmoving, and another circle that is breathing—expanding and contracting—right before your eyes. Our goal is to find the exact 'rhythm'—the range of the radius $r$—that forces these two circles to embrace at exactly two distinct points.

Unmasking the First Circle

We begin with the equation of our first circle: $x^2+y^2-10x+16=0$. At first glance, it looks a bit messy, but in mathematics, messiness is just a hidden order waiting to be revealed.

We need to bring this into the standard form of a circle, $(x-h{)}^2+(y-k{)}^2=R^2$, which tells us everything we need to know: the center $(h,k)$ and the radius $R$.

Let's group the $x$ terms: $x^2-10x+y^2+16=0$. To complete the square for $x$, we take half of the coefficient of $x$ (which is $-10$), divide by $2$ to get $-5$, and square it to get $25$.

We add and subtract $25$:

This simplifies beautifully to $(x-5{)}^2+y^2=9$. And there it is! Our first circle is centered at $C_1=(5,0)$ with a radius $r_1=3$.

The Bridge Between Centers

Now, let's look at our second circle: $x^2+y^2=r^2$. This one is much friendlier. It is centered at the origin, $C_2=(0,0)$, and has a radius $r_2=r$.

To understand how these two circles interact, we need to know how far apart they are. The distance $d$ between $C_1(5,0)$ and $C_2(0,0)$ is simply $5$ units. This distance $d=5$ is the bridge that connects our two worlds.

The Geometric Sweet Spot

Here is the core of the problem. For two circles to intersect at exactly two distinct points, they must be in a specific geometric configuration.

The condition for two distinct intersection points is given by the elegant inequality:

Think of this as the 'Goldilocks' zone. The distance $d$ must be greater than the difference of the radii (to avoid internal touching) and less than the sum of the radii (to avoid external separation).

Substituting our values $r_1=3$, $r_2=r$, and $d=5$, we get:

The Algebraic Dance

Now, we solve this compound inequality. We can break it into two manageable pieces.

First, the right side: $5<3+r$. Subtracting $3$ from both sides, we get $r>2$. This is our first constraint.

Second, the left side: $|3-r|<5$. This absolute value inequality expands into:

Subtracting $3$ from all parts gives $-8<-r<2$. Now, we multiply by $-1$. Remember the golden rule of inequalities: when you multiply or divide by a negative number, you must flip the inequality signs!

This transforms our expression into $8>r>-2$. Since a radius cannot be negative, we focus on $r<8$.

The Final Harmony

Combining our two findings, $r>2$ and $r<8$, we arrive at the final range:

If $r$ is exactly $2$ or $8$, the circles kiss at a single point, but they do not intersect at two. By keeping $r$ strictly between $2$ and $8$, we ensure that the circles overlap in that perfect, two-point harmony. You have just solved a classic JEE problem by blending geometric intuition with algebraic precision.