Let the tangent to the circle x² + y² = 25 at the point R(3, 4) meet x -axis and y -axis at point P

Q: Why is the incentre of a right triangle at the origin always $(r_{in}, r_{in})$?

Think about the definition of the incentre. It is the point equidistant from all three sides. Since the sides are the x-axis ($y=0$), the y-axis ($x=0$), and the hypotenuse, the distance to the x-axis is the y-coordinate, and the distance to the y-axis is the x-coordinate. Thus, for the point to be equidistant from both axes, $x$ must equal $y$, which is exactly the in-radius $r_{in}$.

Q: Can I just use the general formula for the incentre $\frac{ax_1+bx_2+cx_3}{a+b+c}$?

Yes, you absolutely can! But in a JEE exam, time is your currency. Using the general formula involves calculating the lengths of all sides and the coordinates of all vertices. The shortcut $r_{in} = \frac{a+b-c}{2}$ is a powerful tool specifically for right triangles. Master the shortcut to save precious minutes.

Q: Why is the radius of the new circle just the distance $OI$?

The problem states the new circle passes through the origin $O$ and has its centre at $I$. By definition, the radius of any circle is the distance from its centre to any point on its circumference. Since the origin lies on the circumference, the distance $OI$ is the radius. It is that simple!

Question

Let the tangent to the circle $x^2 + y^2 = 25$ at the point R(3, 4) meet x -axis and y -axis at point P and Q, respectively. If r is the radius of the circle passing through the origin O and having centre at the incentre of the triangle OPQ, then $r^2$ is equal to :

Options:

(A) $\frac{529}{64}$

(B) $\frac{125}{72}$

(C) $\frac{625}{72}$

(D) $\frac{585}{66}$

Sigma Percentile · Accepted Answer

Welcome, future engineer. Today, we are not just solving a coordinate geometry problem; we are painting a picture of mathematical elegance.

Imagine the circle $x^2 + y^2 = 25$ as a perfect, serene pond centered at the origin. Point $R(3, 4)$ is a pebble dropped on its surface. When we draw the tangent at $R$, we are drawing a line that just kisses the circle. This line, $3x + 4y = 25$, is our gateway. It carves out a triangle $OPQ$ against the axes.

### The Foundation: The Right-Angled Triangle

We have a right-angled triangle $OPQ$. The legs are $OP$ and $OQ$. We find them by setting $x=0$ and $y=0$ in our tangent equation.

We get $P(\frac{25}{3}, 0)$ and $Q(0, \frac{25}{4})$. This is the foundation of our structure. We have a base of $\frac{25}{3}$ and a height of $\frac{25}{4}$.

Before we rush to the incentre, we must find the hypotenuse $PQ$. Using the Pythagorean theorem:

$$PQ = \sqrt{\left(\frac{25}{3}\right)^2 + \left(\frac{25}{4}\right)^2}$$

With a bit of algebraic finesse, we factor out the $25$ and find:

$$PQ = \frac{125}{12}$$

### The Magic of the Incentre

Now, the heart of the problem: the incentre $I$. For a right triangle, the incentre is not just any point; it is the point $(r_{in}, r_{in})$.

This is because the distance from the incentre to the $x$-axis and the $y$-axis must be the same—the in-radius. We use the beautiful identity:

$$r_{in} = \frac{a+b-c}{2}$$

where $a$ and $b$ are the legs and $c$ is the hypotenuse. Substituting our values:

$$r_{in} = \frac{\frac{25}{3} + \frac{25}{4} - \frac{125}{12}}{2}$$

Simplifying this fraction is a test of patience, but it yields $r_{in} = \frac{25}{12}$. Thus, our incentre $I$ is $(\frac{25}{12}, \frac{25}{12})$.

### The Final Stretch

Finally, we are asked for the radius $r$ of a circle centered at $I$ that passes through the origin. The radius is simply the distance $OI$.

Using the distance formula:

$$r = \sqrt{\left(\frac{25}{12}-0\right)^2 + \left(\frac{25}{12}-0\right)^2}$$

This simplifies to $r = \frac{25}{12}\sqrt{2}$. The question asks for $r^2$.

Squaring our result, we get:

$$r^2 = \left(\frac{25}{12}\right)^2 \cdot 2 = \frac{625}{144} \cdot 2 = \frac{625}{72}$$

**Victory!** You have successfully navigated the tangent, the triangle, and the incentre. This is the essence of JEE Advanced—connecting disparate concepts into a single, harmonious solution. The final answer is **$\frac{625}{72}$**.

Select Answer:

Visualized Solution

Visualizing the Circle and Point $R$

Equation of the Tangent at $R (3, 4)$

Finding Intercepts $P$ and $Q$

Analyzing Triangle $O P Q$

Calculating Hypotenuse $P Q$

Formula for In-radius of Right Triangle

Calculating the Incentre $I$

Setting up the Final Radius $r$

Calculating $r^{2}$

Conclusion and Key Takeaway

The Sigma Insight: Equation of Tangent and Normal

The Foundation

The Magic of the Incentre

The Final Stretch

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Select Answer:

Visualized Solution

Visualizing the Circle and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Equation of the Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R(3,4)

Finding Intercepts .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Analyzing Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OPQ

Calculating Hypotenuse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ