Let the tangents drawn from the origin to the circle, x² + y² - 8x - 4y + 16 = 0 touch it at the poi

Q: Wait, why is the tangent length $L = \sqrt{16} = 4$? Shouldn't it be negative since it's a distance?

That is a fantastic question! Remember, the length of a tangent from an external point $(x_1, y_1)$ to a circle $S = 0$ is defined as $\sqrt{S_1}$. Since $S_1$ is the power of the point, and the power of a point outside the circle is always positive, the square root will always yield a positive real value. Distance is always non-negative, so we take the positive root.

Q: Can I just find the coordinates of $A$ and $B$ and use the distance formula? It seems more intuitive.

You absolutely can! You would find the equation of the chord of contact using $T=0$, which is $-4x - 2y + 16 = 0$, or $2x + y = 8$. Then, you would find the intersection of this line with the circle. However, that involves solving a quadratic equation and is prone to calculation errors. Using the formula $AB = \frac{2LR}{\sqrt{L^2 + R^2}}$ is the 'JEE way'—it's faster, more elegant, and significantly reduces the chance of a silly mistake.

Q: Is there a geometric way to derive the chord length formula without memorizing it?

Yes! If you draw the radius $R$ to the point of contact $A$, you get a right triangle $\triangle OAC$ with hypotenuse $OC = d = \sqrt{4^2 + 2^2} = \sqrt{20}$. The altitude from $A$ to $OC$ is half the chord length, let's call it $h = AB/2$. In $\triangle OAC$, the area is $\frac{1}{2} \times R \times L = \frac{1}{2} \times d \times h$. Solving for $h$ gives $h = \frac{RL}{d}$. Since $AB = 2h$, we get $AB = \frac{2RL}{d} = \frac{2RL}{\sqrt{L^2 + R^2}}$. Geometry is beautiful, isn't it?

Question

Let the tangents drawn from the origin to the circle, $x^2 + y^2 - 8x - 4y + 16 = 0$ touch it at the point $A$ and $B$. The $(AB)^2$ is equal to:

Options:

(A) $\frac{64}{5}$

(B) $\frac{24}{5}$

(C) $\frac{8}{5}$

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorer of the mathematical universe! Today, we are going to unravel a classic problem that sits at the heart of coordinate geometry.

We are looking at a circle defined by the equation $x^2 + y^2 - 8x - 4y + 16 = 0$, and we are drawing tangents from the origin $O(0, 0)$ to this circle. Our mission is to find the square of the length of the chord of contact, $AB$.

### Decoding the Circle

Before we can dance with the tangents, we must understand the circle itself. The general form of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.

By comparing our given equation $x^2 + y^2 - 8x - 4y + 16 = 0$ to this general form, we extract the vital signs: $g = -4$, $f = -2$, and $c = 16$.

With these, we locate the center $C(-g, -f)$ at $(4, 2)$. The radius $R$ is calculated as follows:

$$R = \sqrt{g^2 + f^2 - c} = \sqrt{(-4)^2 + (-2)^2 - 16} = \sqrt{16 + 4 - 16} = 2$$

Our circle is centered at $(4, 2)$ with a radius of $2$.

### The Power of the Point

Now, let's turn our attention to the origin $O(0, 0)$. When we draw tangents from an external point to a circle, the length of these tangents, $L$, is a fundamental property.

We use the power of the point formula: $L = \sqrt{S_1}$. By substituting the coordinates of the origin $(0, 0)$ into the circle's equation, we get:

$$L = \sqrt{0^2 + 0^2 - 8(0) - 4(0) + 16} = \sqrt{16} = 4$$

The length of the tangent from the origin to the points of contact $A$ and $B$ is exactly $4$.

### The Elegant Chord

We could find the coordinates of $A$ and $B$ by finding the intersection of the chord of contact line $T=0$ with the circle, but that is a path filled with algebraic thorns. Instead, we embrace the elegance of the chord of contact formula:

$$AB = \frac{2LR}{\sqrt{L^2 + R^2}}$$

This formula is a direct consequence of the geometric properties of the right-angled triangle formed by the center, the point of contact, and the external point. Substituting $L = 4$ and $R = 2$, we have:

$$AB = \frac{2(4)(2)}{\sqrt{4^2 + 2^2}}$$

### Final Calculation

Simplifying the expression, the numerator becomes $2 	imes 4 	imes 2 = 16$. The denominator is $\sqrt{16 + 4} = \sqrt{20}$.

Thus, $AB = \frac{16}{\sqrt{20}}$. The question asks for the square of the length, $(AB)^2$:

$$(AB)^2 = \left( \frac{16}{\sqrt{20}} ight)^2 = \frac{256}{20}$$

Finally, we reduce this fraction by dividing both the numerator and the denominator by $4$. This leads us to our destination:

$$(AB)^2 = \frac{64}{5}$$

Through careful analysis and the application of geometric principles, we have conquered the problem. Remember, in JEE Advanced, the most complex problems often yield to the most elegant methods.

Animated Solution for Mathematics - Circles: Let the tangents drawn from the origin to the circle, $x^{2} + y^{2} - 8 x - 4 y + 16 = 0$ touch it at the point $A$ and $B$ . The $(A B)^{2}$ is equal to:

Select Answer:

Visualized Solution

The Problem Setup

Circle Equation Analysis

Finding Center and Radius

Length of Tangent Formula

Calculating Tangent Length $L$

Visualizing the Chord of Contact

Chord of Contact Formula

Substituting $L$ and $R$

Simplifying the Denominator

Squaring the Length

Final Fraction Reduction

The Sigma Insight: Length of Tangent and Chord of Contact

Analyzing the Setup

Decoding the Circle

The Power of the Point

The Elegant Chord

Final Calculation

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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; } Let the tangents drawn from the origin to the circle, x2+y2−8x−4y+16=0 touch it at the point A and B. The (AB)2 is equal to: