Let the lines y + 2x = √11 + 7√7 and 2y + x = 2√11 + 6√7 be normal to a circle C : (x - h)² + (y - k

Q: Wait, why do we assume the normals intersect at the center? Is that always true?

That is a fantastic question! Yes, it is always true. By definition, a normal to a circle at any point is a line perpendicular to the tangent at that point. Since the radius is also perpendicular to the tangent at the point of contact, the normal must lie along the line containing the radius. Therefore, every normal passes through the center $(h, k)$.

Q: The tangent equation looks so messy with all those square roots. Is there a trick to simplify it?

I completely understand why that looks intimidating! The trick is not to panic. In JEE Advanced problems, these complex-looking coefficients are often designed to cancel out perfectly when you plug them into the distance formula. Just keep your cool, write out the steps clearly, and trust the algebra. The 'messiness' is usually just a veil covering a very elegant simplification.

Q: Why do we need to calculate $(5h - 8k)^2$ separately? Can't we just find $h$ and $k$ and plug them in?

You absolutely can! Calculating $h$ and $k$ first is the standard, reliable way to do it. However, sometimes calculating the expression $(5h - 8k)$ directly can save you time and reduce the chance of arithmetic errors. Both methods are valid, but always look for the path that feels most comfortable for you.

Question

Let the lines $y + 2x = \sqrt{11} + 7\sqrt{7}$ and $2y + x = 2\sqrt{11} + 6\sqrt{7}$ be normal to a circle $C : (x - h)^2 + (y - k)^2 = r^2$. If the line $\sqrt{11}y - 3x = \frac{5\sqrt{77}}{3} + 11$ is tangent to the circle $C$, then the value of $(5h - 8k)^2 + 5r^2$ is equal to ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler, to the beautiful world of coordinate geometry. Today, we are not just solving a problem; we are uncovering a hidden symmetry.

Imagine a circle, a perfect, balanced entity. We are given two lines, and we are told they are normals. These lines are the guardians of the circle's center, representing the paths that lead directly to the heart of the circle, the point $(h, k)$.

### Finding the Heart of the Circle

We start with two equations representing the normals:
$$y + 2x = \sqrt{11} + 7\sqrt{7}$$
$$2y + x = 2\sqrt{11} + 6\sqrt{7}$$

Because both lines pass through the center $(h, k)$, their intersection is the center itself. We use the method of elimination to solve this system. Multiplying the second equation by $2$, we obtain:
$$4y + 2x = 4\sqrt{11} + 12\sqrt{7}$$

Subtracting the first equation ($y + 2x = \sqrt{11} + 7\sqrt{7}$) from this result, the $2x$ terms vanish. We are left with:
$$3y = 3\sqrt{11} + 5\sqrt{7} \Rightarrow k = y = \sqrt{11} + \frac{5\sqrt{7}}{3}$$

Substituting this value back into the first normal equation, we solve for $h$:
$$h = \frac{8\sqrt{7}}{3}$$

### The Tangent and the Radius

Now, we turn our attention to the tangent line:
$$\sqrt{11}y - 3x = \frac{5\sqrt{77}}{3} + 11$$

This line is a boundary that touches the circle at a single point. The distance from the center $(h, k)$ to this tangent is, by definition, the radius $r$. We use the perpendicular distance formula:
$$r = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$

When we substitute our coordinates $(h, k)$ into the equation, the terms involving $\sqrt{77}$ and the constants simplify significantly. Through algebraic reduction, the expression collapses into a manageable value, leading us to the square of the radius, $r^2 = \frac{448}{20}$.

### Final Calculation

We have our center $(h, k)$ and our radius $r$. The problem asks for the value of $(5h - 8k)^2 + 5r^2$.

First, we calculate the linear component:
$$5h - 8k = -8\sqrt{11}$$
$$(5h - 8k)^2 = (-8\sqrt{11})^2 = 64 	imes 11 = 704$$

Next, we calculate the radial component:
$$5r^2 = 5 	imes \frac{448}{20} = \frac{448}{4} = 112$$

Adding these two results together, we arrive at our final answer:
$$704 + 112 = \mathbf{816}$$

This problem is a testament to the fact that even the most complex-looking expressions in JEE Advanced are often built on a foundation of simple, elegant truths. Keep practicing, keep visualizing, and most importantly, keep falling in love with the process.

Enter Numerical Value:

Visualized Solution

Visualizing the Normals

Aligning Coefficients

Solving for $k$

Solving for $h$

Calculating $(5 h - 8 k)$

Squaring the Result

The Tangent and Radius

Applying Distance Formula

Simplifying the Numerator

Calculating $5 r^{2}$

The Final Answer

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Finding the Heart of the Circle

The Tangent and the Radius

Final Calculation

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