Four distinct points (2k, 3k), (1, 0), (0, 1) and (0, 0) lie on a circle for k equal to : (1) 2/13 (

Q: Why can't we just find the center and radius? It seems like a lot of work to use the diameter form.

You absolutely can! You could find the midpoint of $AB$ to get the center and calculate the distance to find the radius. But in JEE, time is your currency. The diameter form $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$ is a powerful shortcut that bypasses the need to calculate the center and radius explicitly. It’s about working smarter, not just harder.

Q: Wait, why do we reject $k=0$?

Great catch! The problem explicitly states that the four points must be 'distinct'. If $k=0$, the point $P(2k, 3k)$ becomes $(0, 0)$, which is identical to the origin $O$. Since the points cannot overlap, we must discard this solution. Always read the fine print in JEE questions—it’s often where the trap is hidden!

Q: Is there any other way to solve this without the circle equation?

Yes, you could use the property of cyclic quadrilaterals or even the power of a point theorem. However, since we have the coordinates of three points that define the circle uniquely, the circle equation is the most robust and least error-prone method. Stick to the method that gives you the most confidence under pressure.

Question

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

Options:

(A) $\frac{2}{13}$

(B) $\frac{3}{13}$

(C) $\frac{5}{13}$

(D) $\frac{1}{13}$

Sigma Percentile · Accepted Answer

### The Geometry of the Coordinate Plane

Welcome, future engineers. Today, we are going to look at a problem that might seem like a simple exercise in coordinate geometry, but it is actually a masterclass in geometric intuition.

When you see points like $(0, 0)$, $(1, 0)$, and $(0, 1)$, do not just see numbers. See the canvas of the Cartesian plane. See the origin, the x-axis, and the y-axis.

These points are not random; they are the building blocks of a right-angled triangle. When you plot $O(0, 0)$, $A(1, 0)$, and $B(0, 1)$, you are essentially drawing the corner of a square. This is the first step in our journey: recognizing the hidden structure.

### The Right-Angle Insight

Look at the angle $\angle AOB$. Because $A$ lies on the x-axis and $B$ lies on the y-axis, the angle between them is exactly $90^{\circ}$.

This is the 'Aha!' moment. In the world of circles, a right-angled triangle inscribed in a circle is a special guest.

Because of the Thales's Theorem property, the hypotenuse of a right-angled triangle inscribed in a circle is always the diameter of that circle. This realization is your golden ticket. It saves you from the tedious process of finding the center $(h, k)$ and the radius $r$ using the general equation:

$$x^2 + y^2 + 2gx + 2fy + c = 0$$

We can bypass all that.

### The Elegance of the Diameter Form

Since $AB$ is the diameter, we can use the diameter form of the circle equation directly. If a circle has a diameter with endpoints $(x_1, y_1)$ and $(x_2, y_2)$, its equation is given by:

$$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$

Substituting our points $A(1, 0)$ and $B(0, 1)$, we get:

$$(x - 1)(x - 0) + (y - 0)(y - 1) = 0$$

Expanding this, we get $x^2 - x + y^2 - y = 0$, or simply:

$$x^2 + y^2 - x - y = 0$$

This is the equation of our circle. It is clean, it is precise, and it is ready to be tested.

### The Final Challenge: The Variable Point

Now, we introduce the fourth point, $P(2k, 3k)$. The problem tells us that this point lies on the circle.

In the language of mathematics, this means the coordinates of $P$ must satisfy the equation we just derived. So, we substitute $x = 2k$ and $y = 3k$ into our equation:

$$(2k)^2 + (3k)^2 - (2k) - (3k) = 0$$

This is where you must be careful with your algebra. Squaring $2k$ gives $4k^2$, and squaring $3k$ gives $9k^2$. Adding them together, we get $13k^2$.

The linear terms $-2k$ and $-3k$ combine to give $-5k$. Thus, we arrive at the quadratic equation:

$$13k^2 - 5k = 0$$

### The Trap and the Triumph

Factoring this is straightforward:

$$k(13k - 5) = 0$$

This gives us two roots: $k = 0$ and $k = \frac{5}{13}$. Now, pause. This is where many students rush and lose marks.

The question specifies that the four points must be **distinct**. If we choose $k = 0$, our point $P$ becomes $(0, 0)$, which is the origin $O$.

That would mean $P$ and $O$ are the same point, violating the condition of distinctness. Therefore, we must reject $k = 0$.

The only valid solution is **$k = \frac{5}{13}$**. You have navigated the geometry, applied the theorem, solved the algebra, and avoided the trap. That is how you conquer JEE Advanced problems.

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

Select Answer:

Visualized Solution

Identify the Fixed Points

Right-Angled Triangle Property

Diameter of the Circumcircle

Equation of the Circle

Simplify the Circle Equation

Substitute Point $P (2 k, 3 k)$

Form the Quadratic Equation

Solve for $k$

Apply the Distinct Points Constraint

The Sigma Insight: Standard and General Equation of a Circle

The Geometry of the Coordinate Plane

The Right-Angle Insight

The Elegance of the Diameter Form

The Final Challenge

The Trap and the Triumph

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Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Four distinct points (2k,3k),(1,0),(0,1) and (0,0) lie on a circle for k equal to :