Consider a family of circles which are passing through the point (- 1, 1) and are tangent to x-axis.

Q: Wait, why is the radius equal to $|k|$ and not just $k$?

That is a fantastic observation! The radius of a circle is a distance, and distance must always be non-negative. If the center of the circle is below the x-axis, the y-coordinate $k$ would be negative. However, the distance from $(h, k)$ to the x-axis is $|k|$. By using the absolute value, we ensure our radius formula holds true regardless of whether the circle is above or below the x-axis.

Q: Why did we square both sides? Can't we just solve for $k$ directly?

Squaring both sides is a standard algebraic maneuver to eliminate the square root and the absolute value. If we didn't square, we would be left with a radical expression like $\sqrt{(h+1)^2 + (k-1)^2} = |k|$, which is much harder to manipulate. Squaring turns the equation into a simple polynomial form, allowing us to see the relationship between $h$ and $k$ much more clearly.

Q: How do we know $(h+1)^2 \ge 0$ is the key? What if $h$ is a complex number?

In the context of coordinate geometry for JEE, we are working within the real coordinate plane. For any real number $h$, the square of $(h+1)$ is always greater than or equal to zero. This is a fundamental property of real numbers. Since our equation equates this non-negative quantity to $2k - 1$, it forces $2k - 1$ to also be non-negative, which is exactly the constraint we need to find the range of $k$.

Question

Consider a family of circles which are passing through the point $(- 1, 1)$ and are tangent to x-axis. If $(h, k)$ are the coordinate of the centre of the circles, then the set of values of $k$ is given by the interval

Options:

(A) $-\frac{1}{2} \le k \le \frac{1}{2}$

(B) $k \le \frac{1}{2}$

(C) $0 \le k \le \frac{1}{2}$

(D) $k \ge \frac{1}{2}$

Sigma Percentile · Accepted Answer

### The Dance of the Circles: A Geometric Journey

Imagine you are standing on the Cartesian plane. You have a fixed anchor point, $P(-1, 1)$, and you are tasked with drawing a family of circles.

But there is a catch: every single circle you draw must be perfectly tangent to the x-axis. As you vary the size of these circles, their centers shift, tracing out a path. Today, we are going to uncover the hidden constraint that governs the y-coordinate of these centers.

### Phase 1: The Geometry of Tangency

Let's place the center of one such circle at $C(h, k)$. If this circle is tangent to the x-axis, what does that tell us?

If you drop a perpendicular line from the center $C$ down to the x-axis, that line segment is the radius $r$. Since the y-coordinate of the center is $k$, the length of this segment is simply the absolute value of $k$.

Thus, we establish our first vital relationship: $r = |k|$. This is the geometric soul of our problem—the radius is locked to the vertical position of the center.

### Phase 2: The Distance Constraint

Now, consider the anchor point $P(-1, 1)$. We know this point lies on the circumference of every circle in our family.

By the definition of a circle, the distance from the center $C(h, k)$ to any point on the circle must be equal to the radius $r$. So, the distance $CP$ must also be $r$. Using the distance formula, we write:

$$\sqrt{(h - (-1))^2 + (k - 1)^2} = r$$

Substituting our previous finding, $r = |k|$, we get:

$$\sqrt{(h + 1)^2 + (k - 1)^2} = |k|$$

### Phase 3: The Algebraic Dance

This equation looks a bit intimidating with the square root and the modulus, but let's simplify it. We square both sides to clear the radical and the absolute value:

$$(h + 1)^2 + (k - 1)^2 = k^2$$

Now, let's expand the term $(k - 1)^2$ using the identity $(a - b)^2 = a^2 - 2ab + b^2$:

$$(h + 1)^2 + (k^2 - 2k + 1) = k^2$$

Look at the beauty of this simplification! We have a $k^2$ on both sides of the equation. Subtracting $k^2$ from both sides, they vanish, leaving us with a much cleaner expression:

$$(h + 1)^2 - 2k + 1 = 0$$

Rearranging to isolate the perfect square, we get:

$$(h + 1)^2 = 2k - 1$$

### Phase 4: The "Aha!" Moment

This is where the magic happens. We have a perfect square, $(h + 1)^2$, on the left side. We know that for any real coordinate $h$, the square of a real number is always non-negative.

That is, $(h + 1)^2 \ge 0$. Because the left side is non-negative, the right side, $2k - 1$, must also be non-negative to maintain the equality.

This gives us the inequality:

$$2k - 1 \ge 0$$

Solving for $k$ is straightforward: $2k \ge 1$, which simplifies to:

$$k \ge \frac{1}{2}$$

### Conclusion

And there we have it! The y-coordinate of the center of any such circle must be at least $\frac{1}{2}$.

Geometrically, this means that no circle in this family can have its center below the horizontal line $y = \frac{1}{2}$. We have successfully translated a geometric constraint into an algebraic inequality, revealing the hidden boundaries of our family of circles.

**The final constraint is $k \ge \frac{1}{2}$.**

Animated Solution for Mathematics - Circles: Consider a family of circles which are passing through the point $(- 1, 1)$ and are tangent to x-axis. If $(h, k)$ are the coordinate of the centre of the circles, then the set of values of $k$ is given by the interval

Select Answer:

Visualized Solution

The Given Point

The Family of Circles

Tangency Condition

Distance to Point $P$

Equating the Distances

Squaring Both Sides

Expanding the Terms

Simplifying the Equation

Isolating the Perfect Square

The Non-Negative Property

Establishing the Inequality

Final Range of $k$

The Sigma Insight: Standard and General Equation of a Circle

The Dance of the Circles

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

Similar Questions

Four distinct points $(2 k, 3 k), (1, 0), (0, 1)$ and $(0, 0)$ lie on a circle for $k$ equal to :

A circle is given by $x^{2} + (y - 1)^{2} = 1$ , another circle $C$ touches it externally and also the x-axis, then the locus of its centre is

Let the centre of a circle, passing through the points $(0, 0), (1, 0)$ and touching the circle $x^{2} + y^{2} = 9$ , be $(h, k)$ . Then for all possible values of the coordinates of the centre $(h, k)$ , $4 (h^{2} + k^{2})$ is equal to

If the two circles $(x - 1)^{2} + (y - 3)^{2} = r^{2}$ and $x^{2} + y^{2} - 8 x + 2 y + 8 = 0$ intersect in two distinct points, then

The locus of the centres of the circles, which touch the circle, $x^{2} + y^{2} = 1$ externally, also touch the y-axis and lie in the first quadrant, is :

The centres of a set of circles, each of radius 3, lie on the circle $x^{2} + y^{2} = 25$ . The locus of any point in the set is

The circle passing through the point $(- 1, 0)$ and touching the y-axis at $(0, 2)$ also passes through the point

Let the equation of the circle, which touches x-axis at the point $(a, 0)$ , $a > 0$ and cuts off an intercept of length b on y-axis be $x^{2} + y^{2} - α x + β y + γ = 0$ If the circle lies below x-axis, then the ordered pair $(2 a, b^{2})$ is equal to

If the two circles $(x - 1)^{2} + (y - 3)^{2} = r^{2}$ and $x^{2} + y^{2} - 8 x + 2 y + 8 = 0$ intersect in two distinct point, then