For the circle x² + y² = r², find the value of r for which the area enclosed by the tangents drawn f

Q: Wait, why is the area formula $A = \frac{r L^3}{r^2 + L^2}$? Can we derive it?

That is a fantastic question! The triangle is composed of two right-angled triangles. The area is $r \times L \times \sin(\theta)$, where $\theta$ is the angle between the tangent and the line connecting the center to $P$. By using the properties of the right triangle formed by the radius, tangent, and the line to $P$, we can express $\sin(\theta)$ in terms of $r$ and $L$, leading to this elegant formula. It is a powerful tool to have in your JEE toolkit!

Q: Why can we ignore the denominator when differentiating?

Because the denominator $r^2 + L^2$ simplifies to $D^2$, which is a constant $100$ in this problem. When you set the derivative of a function $\frac{f(r)}{C}$ to zero, the constant $C$ does not affect the location of the critical point. It makes the calculus much cleaner!

Q: Why must $r < 10$?

Great observation! If $r \ge 10$, the point $P$ would be inside or on the circle, and tangents cannot be drawn from an internal point. The geometry of the problem requires $P$ to be external, so $r$ must be strictly less than the distance $D=10$.

Question

For the circle $x^2 + y^2 = r^2$, find the value of $r$ for which the area enclosed by the tangents drawn from the point $P(6, 8)$ to the circle and the chord of contact is maximum.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at a circle centered at the origin with an unknown radius $r$. You have a point $P$ sitting at $(6, 8)$.

The distance from the origin to this point is $D = \sqrt{6^2 + 8^2} = 10$. This is our fixed anchor.

Now, imagine drawing two tangents from $P$ to the circle. These tangents, along with the chord of contact, form a triangle. Our goal is to find the radius $r$ that makes this triangle as large as possible.

### The Foundation

First, we need the length of the tangent $L$. From the properties of circles, we know $L = \sqrt{D^2 - r^2}$. Substituting our known distance $D=10$, we get $L = \sqrt{100 - r^2}$.

The area of the triangle formed by the tangents and the chord of contact is given by the formula:

$$A = \frac{r L^3}{r^2 + L^2}$$

Notice the denominator $r^2 + L^2$. Since $L^2 = 100 - r^2$, the denominator becomes $r^2 + (100 - r^2) = 100$, which is simply $D^2$.

Thus, our area function simplifies to:

$$A(r) = \frac{r (100 - r^2)^{\frac{3}{2}}}{100}$$

### The Calculus of Optimization

To maximize the area, we need to find where the derivative $\frac{dA}{dr}$ is zero. Since $100$ is a constant, we focus on the numerator $f(r) = r(100 - r^2)^{\frac{3}{2}}$.

Applying the product rule $\frac{d}{dr}(u \cdot v) = u'v + uv'$, we obtain:

$$1 \cdot (100 - r^2)^{\frac{3}{2}} + r \cdot \frac{3}{2}(100 - r^2)^{\frac{1}{2}}(-2r) = 0$$

Simplifying this, the factor of $2$ cancels out, leaving us with:

$$(100 - r^2)^{\frac{3}{2}} - 3r^2(100 - r^2)^{\frac{1}{2}} = 0$$

Factoring out $(100 - r^2)^{\frac{1}{2}}$, we get:

$$(100 - r^2)^{\frac{1}{2}} [ (100 - r^2) - 3r^2 ] = 0$$

Since $r < 10$, the first term cannot be zero. Thus, we must have $100 - 4r^2 = 0$.

This leads us to $4r^2 = 100$, or $r^2 = 25$, which gives **$r = 5$**.

### The Elegant Shortcut

We have arrived at our answer: **$r = 5$**. Note that the distance $D$ was $10$, and our optimal radius $r$ is $5$, which is exactly $D/2$.

This is a universal truth for this configuration: the area of the triangle formed by the tangents and the chord of contact is always maximized when the radius is half the distance of the external point from the center. Keep this in your arsenal to solve similar problems efficiently.

Animated Solution for Mathematics - Circles: For the circle $x^{2} + y^{2} = r^{2}$ , find the value of $r$ for which the area enclosed by the tangents drawn from the point $P (6, 8)$ to the circle and the chord of contact is maximum.

Enter Numerical Value:

Visualized Solution

Visualizing the Geometry

Length of the Tangent $L$

Area of the Triangle

Simplifying the Area Function

Area as a Function of $r$

Differentiating for Maxima

Applying the Product Rule

Factoring the Derivative

Solving for $r$

Key Takeaway

The Sigma Insight: Length of Tangent and Chord of Contact

Analyzing the Setup

The Foundation

The Calculus of Optimization

The Elegant Shortcut

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