How do I determine whether a line intersects, touches, or misses a circle without solving quadratic equations?

Calculate the perpendicular distance $p$ from the center $(h,k)$ to the line $Ax + By + C = 0$ using $p = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}}$. Compare $p$ with radius $r$: if $p r$, the line lies completely outside the circle.

When should I prefer the parametric form $(h + r\cos\theta, k + r\sin\theta)$ over Cartesian coordinates?

Use parametric form whenever a problem involves maximizing or minimizing expressions depending on points on a circle, finding locus of points at specified angular positions, or dealing with symmetric distance problems. It converts a two-variable algebraic system into a single angular variable $\theta$, simplifying optimization immensely.

Conic Sections - Circles JEE Notes, Formulas, Traps & PYQs | ExamiFly

Beta

Please wait while we prepare everything for you

Conic Sections - Circles JEE Notes, Formulas, Traps & PYQs | ExamiFly | Examifly

Solved High-Yield Sample Questions

Practice Mode

Question #1JEE Main 2024

Find the radius of the circle which passes through the origin $(0,0)$ and cuts the circles $x^2 + y^2 - 4x + 3 = 0$ and $x^2 + y^2 - 2y + 1 = 0$ orthogonally.

Solution Locked

Step-by-Step Solution & Concept

Step 1

Let the required circle be $x^2 + y^2 + 2gx + 2fy + c = 0$ . Since it passes through $(0,0)$ , substitution yields $c = 0$ .

Step 2

Apply the orthogonality condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ . For $x^2 + y^2 - 4x + 3 = 0$ , we have $g_1 = -2, f_1 = 0, c_1 = 3$ , giving $2g(-2) + 0 = 0 + 3 \implies g = -\frac{3}{4}$ . For $x^2 + y^2 - 2y + 1 = 0$ , $g_2 = 0, f_2 = -1, c_2 = 1$ , giving $0 + 2f(-1) = 0 + 1 \implies f = -\frac{1}{2}$ .

Step 3

Calculate radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{\left(-\frac{3}{4}\right)^2 + \left(-\frac{1}{2}\right)^2 - 0} = \sqrt{\frac{9}{16} + \frac{1}{4}} = \frac{\sqrt{13}}{4}$ .

Select an option & click Check Answer to reveal solution

Question #2JEE Main 2023

The line $y = mx + 1$ intersects the circle $x^2 + y^2 = 1$ at two distinct points $A$ and $B$ . If the length of the chord $AB$ is $\sqrt{2}$ , then the value of $m$ is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

For the circle $x^2 + y^2 = 1$ , center is $C(0,0)$ and radius $R = 1$ . The perpendicular distance $p$ from $C(0,0)$ to the line $mx - y + 1 = 0$ is $p = \frac{|0 - 0 + 1|}{\sqrt{m^2 + 1}} = \frac{1}{\sqrt{m^2 + 1}}$ .

Step 2

Relate chord length $L = 2\sqrt{R^2 - p^2}$ . Given $L = \sqrt{2}$ , we have $2\sqrt{1 - p^2} = \sqrt{2} \implies \sqrt{1 - p^2} = \frac{1}{\sqrt{2}} \implies 1 - p^2 = \frac{1}{2} \implies p = \frac{1}{\sqrt{2}}$ .

Step 3

Equate $p = \frac{1}{\sqrt{m^2 + 1}} = \frac{1}{\sqrt{2}} \implies m^2 + 1 = 2 \implies m^2 = 1 \implies m = \pm 1$ .

Select an option & click Check Answer to reveal solution

If you want to explore more chapters of Mathematics then:

Click Here

Mastering Conic Sections - Circles: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

General Equation of a Circle

Equation of Tangent in Slope Form

Equation of Tangent at Point $(x_1, y_1)$

Length of Tangent from External Point

Condition for Orthogonality of Two Circles

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

Master Conic Sections - Circles with 250+ Curated JEE Main & Advanced Drills & AI Error-Correction