Find the equation of the common tangent in 1st quadrant to the circle x² + y² = 16 and the ellipse x

Q: Why do we use the formula $y = mx + \sqrt{a^2m^2 + b^2}$ instead of solving the quadratic equation by substituting the line into the ellipse?

Great question! While substituting the line into the ellipse and setting the discriminant to zero is mathematically valid, it is computationally heavy and prone to algebraic errors. The formula $c^2 = a^2m^2 + b^2$ is a derived shortcut specifically for ellipses. In JEE Advanced, efficiency is key—use the tools that save you time and mental energy.

Q: Why did we choose the negative slope $m = -\frac{2}{\sqrt{3}}$?

Look at the geometry! We are in the first quadrant. For a line to be tangent to both the circle and the ellipse in the first quadrant, it must connect the positive x-axis and the positive y-axis. A line that connects two positive intercepts must have a negative slope (it's falling from left to right). If you chose the positive slope, you would be looking at a tangent in a different quadrant.

Q: Is there a faster way to find the intercept length without finding the full equation?

You could potentially use the property that for a tangent to an ellipse, the product of the intercepts is related to the axes, but finding the full equation is safer and less prone to conceptual traps. Stick to the method of finding the line equation first—it builds a robust understanding of the geometry.

Question

Find the equation of the common tangent in 1st quadrant to the circle $x^2 + y^2 = 16$ and the ellipse $\frac{x^2}{25} + \frac{y^2}{4} = 1$. Also find the length of the intercept of the tangent between the coordinate axes.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing at the origin of a coordinate plane. To your right, an ellipse stretches out, its semi-major axis reaching $a=5$ units along the $x$-axis and its semi-minor axis $b=2$ units along the $y$-axis. Nestled within this, a circle of radius $r=4$ sits perfectly centered at the origin.

Our mission is to find the line that is tangent to both curves in the first quadrant. This is a dance of geometry where we must reconcile the properties of two distinct conic sections.

### The Master Equation

We start with the general equation of a tangent to an ellipse:
$$y = mx + \sqrt{a^2m^2 + b^2}$$

For our specific ellipse, $a^2 = 25$ and $b^2 = 4$. Substituting these values, our line equation becomes:
$$y = mx + \sqrt{25m^2 + 4}$$

Now, we pivot to the circle $x^2 + y^2 = 16$. A line is tangent to a circle if and only if the perpendicular distance from the center $(0,0)$ to the line is exactly equal to the radius, $r=4$.

### Solving for the Slope

Using the perpendicular distance formula from the origin to the line $mx - y + \sqrt{25m^2 + 4} = 0$, we have:
$$\frac{|\sqrt{25m^2 + 4}|}{\sqrt{m^2 + 1}} = 4$$

Squaring both sides leads to our moment of truth:
$$\frac{25m^2 + 4}{m^2 + 1} = 16$$

Solving this algebraic expression:
$$25m^2 + 4 = 16m^2 + 16$$
$$9m^2 = 12$$
$$m^2 = \frac{4}{3}$$

Since we are working in the first quadrant and the tangent must bridge the curves, the slope must be negative:
$$m = -\frac{2}{\sqrt{3}}$$

### Final Calculation

Substituting the slope back into our tangent equation, we find:
$$y = -\frac{2}{\sqrt{3}}x + \sqrt{25\left(\frac{4}{3}ight) + 4}$$
$$y = -\frac{2}{\sqrt{3}}x + \sqrt{\frac{100+12}{3}} = -\frac{2}{\sqrt{3}}x + \sqrt{\frac{112}{3}}$$
$$y = -\frac{2}{\sqrt{3}}x + 4\sqrt{\frac{7}{3}}$$

To find the intercepts, we set $x=0$ to find the $y$-intercept $(0, 4\sqrt{7/3})$ and $y=0$ to find the $x$-intercept $(2\sqrt{7}, 0)$. Calculating the distance between these points, we arrive at the final result:
$$	ext{Distance} = \frac{14}{\sqrt{3}}$$

### The Philosophy of the Tangent

When you approach a problem involving multiple conic sections, the biggest mistake is to treat them as separate entities. They are linked by the shared line—the tangent.

By expressing the tangent in terms of the ellipse's parameters, you have already 'baked in' the condition that the line touches the ellipse. The circle then acts as a filter, forcing the slope $m$ to take a specific value.

This is the essence of **JEE Advanced problem solving**: using one constraint to define the variable, and the second to solve for it. Trust the process, trust your derivation, and the numbers will always fall into place.

Animated Solution for Mathematics - Conic Sections: Find the equation of the common tangent in 1st quadrant to the circle $x^{2} + y^{2} = 16$ and the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{4} = 1$ . Also find the length of the intercept of the tangent between the coordinate axes.

Visualized Solution

Visualizing the Curves

General Tangent to an Ellipse

Substituting Ellipse Parameters

Condition for Circle Tangency

Applying the Distance Formula

Squaring Both Sides

Cross-Multiplying and Simplifying

Solving for $m^{2}$

Determining the Slope $m$

The Common Tangent Equation

Finding the Intercepts

Calculating Intercept Length

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Master Equation

Solving for the Slope

Final Calculation

The Philosophy of the Tangent

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