The straight line x + 2y = 1 meets the coordinate axes at A and B. A circle is drawn through A, B an

Q: Wait, why is the line segment AB the diameter? I thought I had to find the center first.

That is a brilliant observation! You don't *have* to find the center. Because the angle between the coordinate axes is $90^\circ$, any triangle formed by the origin and the intercepts $A$ and $B$ is a right-angled triangle. By the property of circles, a right angle subtended at the circumference implies the hypotenuse is the diameter. It saves you so much time!

Q: How did you get the tangent equation $2x + y = 0$ so quickly? I usually use calculus.

Calculus is perfectly valid, but for a circle $x^2 + y^2 + 2gx + 2fy + c = 0$, the tangent at the origin $(0,0)$ is simply the linear part of the equation: $2gx + 2fy = 0$. Since our circle is $x^2 + y^2 - x - \frac{1}{2}y = 0$, the linear part is $-x - \frac{1}{2}y = 0$, which simplifies to $2x + y = 0$. It is a powerful shortcut for JEE!

Q: Why do we take the absolute value in the distance formula?

Distance is a physical length, and length can never be negative. The absolute value ensures that regardless of which side of the line the point lies on, the distance remains a positive magnitude. Always keep those bars there!

Q: Is there a way to solve this without finding the circle equation at all?

You could use coordinate geometry properties, but finding the circle equation is the most robust path here. It anchors your geometry in algebra, preventing errors. Sometimes, the 'long' way is actually the safest way to ensure accuracy under exam pressure.

Q: I keep getting $\frac{5}{2\sqrt{5}}$. How do I simplify that to $\frac{\sqrt{5}}{2}$?

Think of $5$ as $(\sqrt{5})^2$. So, $\frac{(\sqrt{5})^2}{2\sqrt{5}}$ allows you to cancel one $\sqrt{5}$ from the numerator and denominator, leaving you with $\frac{\sqrt{5}}{2}$. It is just basic surd manipulation!

Question

The straight line $x + 2y = 1$ meets the coordinate axes at A and B. A circle is drawn through A, B and the origin. Then the sum of perpendicular distances from A and B on the tangent to the circle at the origin is :

Options:

(A) $\frac{\sqrt{5}}{4}$

(B) $\frac{\sqrt{5}}{2}$

(C) $2\sqrt{5}$

(D) $4\sqrt{5}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE landscape. Today, we are not just solving a problem; we are uncovering the hidden symmetry of a circle dancing between the axes.

Imagine you are standing on a Cartesian plane. You see a line, $x + 2y = 1$, cutting through the space, creating two distinct points of intersection with our axes. These points, $A$ and $B$, are the anchors of our story.

### Finding the Anchors

First, let us locate these points. When the line kisses the x-axis, the y-coordinate vanishes, leaving us with $x = 1$. Thus, $A$ is at $(1, 0)$.

When it kisses the y-axis, the x-coordinate vanishes, and $2y = 1$, giving us $y = \frac{1}{2}$. So, $B$ is at $(0, \frac{1}{2})$. We now have our two points, $A(1, 0)$ and $B(0, \frac{1}{2})$, and we know the circle must also pass through the origin $O(0, 0)$.

### The Diameter Revelation

Here is where the magic happens. Because the coordinate axes are perpendicular, the angle $\angle AOB$ is exactly $90^{\circ}$.

In the world of geometry, if a circle passes through the vertices of a right-angled triangle, the hypotenuse must be the diameter. Therefore, the segment $AB$ is the diameter of our circle.

Using the diameter form of a circle equation, $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$, we substitute our points to get $(x - 1)(x - 0) + (y - 0)(y - \frac{1}{2}) = 0$. Expanding this, we arrive at the beautiful equation:

$$x^2 + y^2 - x - \frac{1}{2}y = 0$$

### The Tangent at the Origin

Now, we seek the tangent to this circle at the origin. While you could use derivatives, there is a more elegant path.

For any circle $x^2 + y^2 + 2gx + 2fy + c = 0$, the tangent at the origin is simply the linear part of the equation: $2gx + 2fy = 0$. In our case, this gives us $-x - \frac{1}{2}y = 0$.

Multiplying by $-2$ to clean it up, we find our tangent line:

$$2x + y = 0$$

### The Final Calculation

We are almost there. We need the sum of the perpendicular distances from $A(1, 0)$ and $B(0, \frac{1}{2})$ to the line $2x + y = 0$. Using the distance formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$, we calculate:

For point $A(1, 0)$:

$$d_1 = \frac{|2(1) + 0|}{\sqrt{2^2 + 1^2}} = \frac{2}{\sqrt{5}}$$

For point $B(0, \frac{1}{2})$:

$$d_2 = \frac{|2(0) + \frac{1}{2}|}{\sqrt{2^2 + 1^2}} = \frac{1/2}{\sqrt{5}} = \frac{1}{2\sqrt{5}}$$

Summing these distances, we get $\frac{2}{\sqrt{5}} + \frac{1}{2\sqrt{5}} = \frac{4 + 1}{2\sqrt{5}} = \frac{5}{2\sqrt{5}}$. By rationalizing the numerator, we see that:

$$\frac{5}{2\sqrt{5}} = \frac{\sqrt{5} \cdot \sqrt{5}}{2\sqrt{5}} = \frac{\sqrt{5}}{2}$$

And there it is! The complexity collapses into a simple, elegant result. The final answer is **$\frac{\sqrt{5}}{2}$**. Remember, every time you face a problem like this, look for the geometric shortcuts first. They are the keys to mastering the JEE.

Animated Solution for Mathematics - Circles: The straight line $x + 2 y = 1$ meets the coordinate axes at A and B. A circle is drawn through A, B and the origin. Then the sum of perpendicular distances from A and B on the tangent to the circle at the origin is :

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Visualized Solution

The Given Line $x + 2 y = 1$

Finding Point $A$

Finding Point $B$

The Circle Through $O, A, B$

Equation of the Circle

Tangent at the Origin

Distance Formula Setup

Distance from Point $A$

Distance from Point $B$

Summing the Distances

Final Simplification

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Finding the Anchors

The Diameter Revelation

The Tangent at the Origin

The Final Calculation

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Visualized Solution

The Given Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x+2y=1

Finding Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Finding Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

The Circle Through .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } O,A,B