How to Solve: Circle (General Equation, Tangent, Normal, Chord of Contact, Director Circle, Radical Axis) – IIT JEE Guide

How to Solve: Circle (General Equation, Tangent, Normal, Chord of Contact, Director Circle, Radical Axis) – IIT JEE Guide

Introduction

Mastering circles in coordinate geometry unlocks 10-15 marks in IIT JEE (Main + Advanced) – enough to push you from a 90th to a 99th percentile rank. Whether it’s finding the equation of a tangent, proving two circles are orthogonal, or solving radical axis problems, this topic appears every year in both papers. If you can solve circle problems fast and error-free, you gain a huge time advantage in the exam.

What You Need To Know First

Before diving in, ensure you’re 100% clear on:
1. Equation of a line (slope-intercept, point-slope, two-point form).
2. Distance between two points and distance from a point to a line.
3. Basic circle properties (radius, center, chord, tangent).

If any of these feel shaky, stop now and revise them first.

Key Vocabulary

Formulas To Know

1. General Equation of a Circle

[ x^2 + y^2 + 2gx + 2fy + c = 0 ] - Center: ((-g, -f)) - Radius: (r = \sqrt{g^2 + f^2 - c}) - Condition for real circle: (g^2 + f^2 - c > 0) - MEMORISE THIS (Not given in JEE sheet)

2. Equation of Tangent to a Circle

(a) At a point ((x_1, y_1)) on the circle

For circle (x^2 + y^2 + 2gx + 2fy + c = 0): [ xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0 ] - MEMORISE THIS (Not given in JEE sheet)

For circle (x^2 + y^2 = r^2) (simplified form): [ xx_1 + yy_1 = r^2 ]

(b) From an external point ((x_1, y_1))

• Length of tangent: (L = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c})
• Equation of tangent: Use pair of tangents formula (see below).
• MEMORISE LENGTH FORMULA (Not given in JEE sheet)

(c) Pair of Tangents from ((x_1, y_1))

[ (x^2 + y^2 + 2gx + 2fy + c)(x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c) = (xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c)^2 ] - Given on exam sheet (But practice deriving it!)

3. Equation of Normal

• Slope of tangent at ((x_1, y_1)): (m_T = -\frac{x_1 + g}{y_1 + f})
• Slope of normal: (m_N = \frac{y_1 + f}{x_1 + g}) (negative reciprocal)
• Equation: (y - y_1 = m_N (x - x_1))

4. Chord of Contact

For circle (x^2 + y^2 + 2gx + 2fy + c = 0) and external point ((x_1, y_1)): [ xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0 ] - MEMORISE THIS (Not given in JEE sheet)

For circle (x^2 + y^2 = r^2): [ xx_1 + yy_1 = r^2 ]

5. Director Circle

For circle (x^2 + y^2 + 2gx + 2fy + c = 0): [ x^2 + y^2 + 2gx + 2fy + (c + r^2) = 0 ] - For (x^2 + y^2 = r^2): Director circle is (x^2 + y^2 = 2r^2). - MEMORISE THIS (Not given in JEE sheet)

6. Radical Axis

For two circles: [ S_1: x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0 ] [ S_2: x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0 ] Radical axis: (S_1 - S_2 = 0) [ 2(g_1 - g_2)x + 2(f_1 - f_2)y + (c_1 - c_2) = 0 ] - Given on exam sheet (But practice deriving it!)

Step-by-Step Method

How to Solve Any Circle Problem (IIT JEE)

Follow these 5 steps for every circle problem:

1. Identify the circle’s equation (general or standard form).
2. Find center and radius (if not given).
3. Determine what’s asked (tangent, normal, chord, etc.).
4. Apply the correct formula (from the list above).
5. Simplify and verify (check if the answer makes sense).

Worked Example (Using Steps)

Problem: Find the equation of the tangent to the circle (x^2 + y^2 - 4x + 6y - 3 = 0) at the point ((5, -2)).

Step 1: Identify the circle’s equation. - Given: (x^2 + y^2 - 4x + 6y - 3 = 0)

Step 2: Find center and radius. - Compare with (x^2 + y^2 + 2gx + 2fy + c = 0): - (2g = -4 \implies g = -2) - (2f = 6 \implies f = 3) - (c = -3) - Center: ((-g, -f) = (2, -3)) - Radius: (r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 9 + 3} = \sqrt{16} = 4)

Step 3: Determine what’s asked. - Tangent at point ((5, -2)).

Step 4: Apply the formula. - Use tangent at ((x_1, y_1)): [ xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0 ] - Plug in ((x_1, y_1) = (5, -2)), (g = -2), (f = 3), (c = -3): [ x(5) + y(-2) + (-2)(x + 5) + 3(y - 2) - 3 = 0 ] [ 5x - 2y - 2x - 10 + 3y - 6 - 3 = 0 ] [ 3x + y - 19 = 0 ]

Step 5: Simplify and verify. - Final equation: (3x + y - 19 = 0) - Check: Does ((5, -2)) lie on the tangent? (3(5) + (-2) - 19 = 15 - 2 - 19 = -6 \neq 0) → Oops! - Mistake spotted: Forgot to use the correct tangent formula for general circle. - Correct approach: Use (xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0) properly. - Recalculate: [ 5x - 2y - 2(x + 5) + 3(y - 2) - 3 = 0 ] [ 5x - 2y - 2x - 10 + 3y - 6 - 3 = 0 ] [ 3x + y - 19 = 0 ] - Verification: Plug ((5, -2)) → (15 - 2 - 19 = -6 \neq 0) → Still wrong! - Realization: The point ((5, -2)) must lie on the circle for the tangent formula to work. - Check if ((5, -2)) lies on the circle: [ 5^2 + (-2)^2 - 4(5) + 6(-2) - 3 = 25 + 4 - 20 - 12 - 3 = -6 \neq 0 ] - Conclusion: The point is not on the circle → No tangent exists at this point. - Exam trap: Always verify the point lies on the circle before finding the tangent.

Worked Examples

Example 1 – Basic (Tangent at a Point)

Problem: Find the equation of the tangent to (x^2 + y^2 = 25) at ((3, 4)).

Solution:
1. Circle: (x^2 + y^2 = 25) (center ((0,0)), radius (5)).
2. Point ((3,4)) lies on the circle (since (3^2 + 4^2 = 25)).
3. Use tangent formula: (xx_1 + yy_1 = r^2).
4. Plug in ((3,4)): (3x + 4y = 25).

Answer: (3x + 4y = 25)

What we did and why: - Used the simplified tangent formula for (x^2 + y^2 = r^2). - Verified the point lies on the circle first.

Example 2 – Medium (Chord of Contact)

Problem: Find the chord of contact from ((5, 3)) to the circle (x^2 + y^2 - 4x - 6y + 9 = 0).

Solution:
1. Circle: (x^2 + y^2 - 4x - 6y + 9 = 0) (general form).
2. Compare with (x^2 + y^2 + 2gx + 2fy + c = 0): - (2g = -4 \implies g = -2) - (2f = -6 \implies f = -3) - (c = 9)
3. Chord of contact formula: (xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0).
4. Plug in ((5,3)): [ 5x + 3y + (-2)(x + 5) + (-3)(y + 3) + 9 = 0 ] [ 5x + 3y - 2x - 10 - 3y - 9 + 9 = 0 ] [ 3x - 10 = 0 \implies x = \frac{10}{3} ]

Answer: (x = \frac{10}{3})

What we did and why: - Used the general chord of contact formula. - Simplified carefully to avoid sign errors.

Example 3 – Exam-Style (Radical Axis)

Problem: Find the radical axis of the circles: [ S_1: x^2 + y^2 - 4x - 6y + 9 = 0 ] [ S_2: x^2 + y^2 - 8x - 2y + 16 = 0 ]

Solution:
1. Radical axis formula: (S_1 - S_2 = 0).
2. Subtract (S_2) from (S_1): [ (x^2 + y^2 - 4x - 6y + 9) - (x^2 + y^2 - 8x - 2y + 16) = 0 ] [ -4x - 6y + 9 + 8x + 2y - 16 = 0 ] [ 4x - 4y - 7 = 0 \implies 4x - 4y = 7 ]

Answer: (4x - 4y = 7)

What we did and why: - Used (S_1 - S_2) to find the radical axis. - Simplified to get the linear equation of the radical axis.

Common Mistakes

Exam Traps

1-Minute Recap

Listen up! Here’s what you must remember for IIT JEE circles:

1. General equation: (x^2 + y^2 + 2gx + 2fy + c = 0) → Center ((-g, -f)), radius (\sqrt{g^2 + f^2 - c}).
2. Tangent at ((x_1, y_1)): Use (xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0).
3. Chord of contact: Same formula as tangent, but for an external point.
4. Director circle: For (x^2 + y^2 = r^2), it’s (x^2 + y^2 = 2r^2).
5. Radical axis: (S_1 - S_2 = 0) → Subtract the two circle equations.
6. Always check:
7. Is the point on the circle? (If not, no tangent exists.)
8. Is the radius real? (If not, the circle doesn’t exist.)
9. Are the circles concentric? (If yes, no radical axis.)

Last tip: In the exam, draw a rough sketch for every problem. It saves time and prevents mistakes.

Now go solve 5 problems in a row without looking at the formulas. That’s how you ace this! ?