Tangent to a Circle

Tangent to a Circle

Hello JEE aspirants! Tangents to circles are fundamental in coordinate geometry and pop up frequently in JEE Main. Mastering this topic will not only boost your score but also build a strong foundation for calculus and other related concepts. Let's dive in!

What is a Tangent?

Geometrically, a tangent to a circle is a line that touches the circle at exactly one point. Imagine a straight road just kissing the edge of a circular park; that road represents a tangent. Understanding this geometric interpretation is crucial for visualizing problems and applying the right formulas.

1. Equation of Tangent at a Point on the Circle

Let's say we have a circle with the equation $x^2 + y^2 + 2gx + 2fy + c = 0$, and we want to find the equation of the tangent at a point $(x_1, y_1)$ lying on the circle. The formula is given by:

Here, $T = 0$ represents the equation of the tangent. Notice how we're replacing $x^2$ with $xx_1$, $y^2$ with $yy_1$, $x$ with $\frac{x + x_1}{2}$, and $y$ with $\frac{y + y_1}{2}$ in the original equation of the circle. This substitution trick works beautifully!

Intuition: Think of it as a weighted average. At the point $(x_1, y_1)$, the tangent's slope is influenced by both $x$ and $x_1$ (and similarly for $y$).

Example: Find the equation of the tangent to the circle $x^2 + y^2 - 4x + 6y - 3 = 0$ at the point $(4, -3)$. Using the formula, we get: $4x - 3y - 2(x + 4) + 3(y - 3) - 3 = 0$, which simplifies to $2x - 8 - 9 - 3 = 0$ or $2x - 20 = 0$, so $x = 10$ and $2x=20$.

2. Tangent in Slope Form

Sometimes, you're given the slope of the tangent and asked to find its equation. For a circle with equation $x^2 + y^2 = r^2$, the tangent with slope $m$ is given by:

Derivation: Let the equation of the tangent be $y = mx + c$. The perpendicular distance from the center $(0, 0)$ to the tangent must be equal to the radius $r$. Using the distance formula: $$\frac{|c|}{\sqrt{1 + m^2}} = r$$ $$c = \pm r\sqrt{1 + m^2}$$ Substituting this back into $y = mx + c$, we get the slope form of the tangent.

Geometric Interpretation: For a given slope $m$, there are two possible tangents to the circle, one on each side. The $\pm$ sign reflects this.

3. Length of Tangent from an External Point

Suppose you have a point $(x_1, y_1)$ outside the circle $x^2 + y^2 + 2gx + 2fy + c = 0$. The length of the tangent from this point to the circle is given by:

Here, $S_1$ represents the value obtained by substituting $(x_1, y_1)$ into the equation of the circle.

Derivation: This formula comes directly from the Pythagorean theorem. If $L$ is the length of the tangent, $R$ is the radius, and $D$ is the distance from $(x_1, y_1)$ to the circle's center, then $L^2 + R^2 = D^2$. Rearranging and using the distance formula gives us the above result.

Geometric Meaning: The length of the tangent represents the distance from the external point to the point where the tangent touches the circle.

Example: From the point $(2,3)$, find the length of the tangent to the circle $x^2 + y^2 + 4x - 6y - 5 = 0$. The length of tangent $= \sqrt{2^2 + 3^2 + 4(2) - 6(3) - 5} = \sqrt{4 + 9 + 8 - 18 - 5} = \sqrt{-2}$, which is not a real number. Did you notice the mistake?

The answer is $\sqrt{-2}$ so this means the point $(2,3)$ is located inside the circle, and no tangent can be found.

4. Pair of Tangents from an External Point

From any point outside a circle, you can draw two tangents. The combined equation of these two tangents is given by:

Where:

• $S = x^2 + y^2 + 2gx + 2fy + c$ (equation of the circle)
• $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$ (value of S at point $(x_1, y_1)$)
• $T = xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c$ (equation of the tangent at a point)

Explanation: This equation represents a second-degree curve that passes through the points where the tangents from $(x_1, y_1)$ touch the circle. It's a compact way to represent both tangent lines simultaneously.

5. Common Tangents to Two Circles

Two circles can have several types of common tangents: direct common tangents (both tangents lie on the same side of the centers) and transverse common tangents (tangents lie on opposite sides of the centers). Finding these tangents involves a bit more algebra and careful consideration of the geometry.

Here are some steps to finding common tangents:

1. Find the centers and radii of both circles. Let them be $(x_1, y_1), r_1$ and $(x_2, y_2), r_2$.
2. Equation of Common Tangent: Assume the equation of the common tangent is $y = mx + c$.
3. Apply Perpendicular Distance Condition: The perpendicular distance from the center of each circle to the tangent must be equal to its radius. This gives you two equations: $$\frac{|mx_1 - y_1 + c|}{\sqrt{m^2 + 1}} = r_1$$ $$\frac{|mx_2 - y_2 + c|}{\sqrt{m^2 + 1}} = r_2$$
4. Solve for $m$ and $c$: Solve these two equations simultaneously to find the values of $m$ and $c$. You will generally get multiple solutions, each corresponding to a different common tangent.

That covers the key concepts about tangents to circles. Practice these formulas, visualize the geometry, and you'll ace those JEE Main problems! Good luck!