Position of Point and Line Relative to Circle

Position of Point and Line Relative to Circle

Hello students! In this lesson, we'll explore the fascinating relationships between points, lines, and circles. This is a crucial topic for JEE Main, forming the foundation for more advanced coordinate geometry problems. Understanding these concepts deeply will not only help you solve problems quickly but also enhance your geometric intuition.

1. Position of a Point Relative to a Circle

Imagine a circle and a point in the same plane. The point can be either inside the circle, on the circle, or outside the circle. But how do we determine this mathematically?

Let's consider a circle with the equation: $$x^2 + y^2 + 2gx + 2fy + c = 0$$

And let $(x_1, y_1)$ be any point in the plane. We define $S_1$ as:

Now, $S_1$ helps us determine the position of the point $(x_1, y_1)$ relative to the circle.

• If $S_1 < 0$, the point $(x_1, y_1)$ lies inside the circle.
• If $S_1 = 0$, the point $(x_1, y_1)$ lies on the circle.
• If $S_1 > 0$, the point $(x_1, y_1)$ lies outside the circle.

Geometric Interpretation: Think of $S_1$ as a "power" value. If you substitute the point's coordinates into the circle's equation, a negative value indicates the point is too close to the center (inside), zero means it's perfectly on the circumference, and a positive value means it's too far (outside).

Example: Consider the circle $x^2 + y^2 = 4$ and the point $(1, 1)$. Here, $S_1 = 1^2 + 1^2 - 4 = -2$. Since $S_1 < 0$, the point $(1, 1)$ lies inside the circle.

2. Line and Circle Intersection

A line can intersect a circle in three possible ways: it can intersect at two distinct points (secant), touch the circle at one point (tangent), or not intersect the circle at all (external).

Consider a line $y = mx + c$ and a circle $x^2 + y^2 = r^2$. To find the points of intersection, substitute the value of $y$ from the line equation into the circle equation:

$$x^2 + (mx + c)^2 = r^2$$

$$x^2 + m^2x^2 + 2mcx + c^2 = r^2$$

$$(1 + m^2)x^2 + 2mcx + (c^2 - r^2) = 0$$

This is a quadratic equation in $x$. Let's analyze the discriminant, $D = b^2 - 4ac$, where $a = (1 + m^2)$, $b = 2mc$, and $c = (c^2 - r^2)$:

• If $D > 0$, the quadratic equation has two distinct real roots, meaning the line intersects the circle at two distinct points (secant).
• If $D = 0$, the quadratic equation has one real root (repeated root), meaning the line touches the circle at one point (tangent).
• If $D < 0$, the quadratic equation has no real roots, meaning the line does not intersect the circle (external).

3. Condition for Tangency

A line is tangent to a circle if it touches the circle at exactly one point. The most common way to express this condition is through the distance from the circle's center to the line.

Let's consider a circle with center $(h, k)$ and radius $r$, and a line $Ax + By + C = 0$. The distance, $d$, from the center to the line is given by:

$$d = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}}$$

For the line to be tangent to the circle, the distance $d$ must be equal to the radius $r$. Therefore, the condition for tangency is:

$$r = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}}$$

Example: Find the condition for the line $y = mx + c$ to be tangent to the circle $x^2 + y^2 = a^2$. The center of the circle is $(0, 0)$ and the radius is $a$. Rewrite the line as $mx - y + c = 0$. Thus,

$$a = \frac{|m(0) - (0) + c|}{\sqrt{m^2 + (-1)^2}} = \frac{|c|}{\sqrt{m^2 + 1}}$$

Squaring both sides, we get: $$a^2 = \frac{c^2}{m^2 + 1}$$ or $$c^2 = a^2(1 + m^2)$$.

4. Length of Chord Cut by Circle on a Line

When a line intersects a circle at two distinct points (secant), it forms a chord. The length of this chord can be calculated using the radius of the circle and the distance of the line from the center.

Where:

• $r$ is the radius of the circle.
• $d$ is the perpendicular distance from the center of the circle to the line.

Derivation: Imagine a right-angled triangle formed by the radius ($r$), half the chord length ($\frac{l}{2}$), and the distance from the center to the line ($d$). By the Pythagorean theorem:

$$r^2 = d^2 + (\frac{l}{2})^2$$

$$(\frac{l}{2})^2 = r^2 - d^2$$

$$\frac{l}{2} = \sqrt{r^2 - d^2}$$

$$l = 2\sqrt{r^2 - d^2}$$

Example: Find the length of the chord cut by the line $x + y = 1$ on the circle $x^2 + y^2 = 1$. The center of the circle is $(0, 0)$ and the radius is $1$. The distance from the center to the line is $d = \frac{|0 + 0 - 1|}{\sqrt{1^2 + 1^2}} = \frac{1}{\sqrt{2}}$. The chord length is therefore $2\sqrt{1^2 - (\frac{1}{\sqrt{2}})^2} = 2\sqrt{1 - \frac{1}{2}} = 2\sqrt{\frac{1}{2}} = \sqrt{2}$.

5. Condition for Line to be Secant, Tangent, or External

This summarizes our previous discussions and provides a single perspective.

Consider a circle with center $C$ and radius $r$, and a line $L$. Let $d$ be the perpendicular distance from the center $C$ to the line $L$. Then:

• If $d < r$, the line $L$ is a secant to the circle (intersects at two points).
• If $d = r$, the line $L$ is a tangent to the circle (touches at one point).
• If $d > r$, the line $L$ is external to the circle (does not intersect).

By mastering these concepts and formulas, you'll be well-equipped to tackle a wide range of JEE Main problems involving circles and lines. Good luck!