The equation of a circle with centre (h, k) and radius r is (x - h)² + (y - k)² = r²
The equation of a circle with centre (0, 0) and radius r is x² + y² = r²
The general equation of a circle is x² + y² + 2gx + 2fy + c = 0
The centre of the circle x² + y² + 2gx + 2fy + c = 0 is (-g, -f)
The radius of the circle x² + y² + 2gx + 2fy + c = 0 is √(g² + f² - c)
The equation of the tangent to a circle at point (x₁, y₁) is xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
The length of the tangent from point (x₁, y₁) to the circle x² + y² + 2gx + 2fy + c = 0 is √(x₁² + y₁² + 2gx₁ + 2fy₁ + c)
The equation of the normal to a circle at point (x₁, y₁) is y - y₁ = -(x₁/(y₁)) × (x - x₁)

🪤 The 5 Mistakes That Cost Marks

Not checking if the given equation represents a circle
Forgetting to find the centre and radius of the circle
Not using the correct equation of the tangent or normal to the circle
Incorrectly finding the length of the tangent from a point to the circle
Not using the correct formula for the equation of the circle ∈ the general form

✏️ 3 Solved PYQs

Question 1: Find the equation of the circle with centre (1, 2) and radius 3 Step 1: Use the equation of a circle with centre (h, k) and radius r Step 2: Substitute h = 1, k = 2, and r = 3 into the equation Step 3: (x - 1)² + (y - 2)² = 3² Step 4: x² - 2x + 1 + y² - 4y + 4 = 9 Step 5: x² + y² - 2x - 4y - 4 = 0
Question 2: Find the centre and radius of the circle x² + y² + 4x + 6y + 4 = 0 Step 1: Compare the given equation with the general equation of a circle Step 2: Find the values of g, f, and c Step 3: g = 2, f = 3, and c = 4 Step 4: Centre = (-g, -f) = (-2, -3) Step 5: Radius = √(g² + f² - c) = √(2² + 3² - 4) = √(4 + 9 - 4) = √9 = 3
Question 3: Find the length of the tangent from point (3, 4) to the circle x² + y² + 2x + 3y - 1 = 0 Step 1: Find the values of g, f, and c Step 2: g = 1, f = (3/2), and c = -1 Step 3: Use the equation of the length of the tangent from point (x₁, y₁) to the circle Step 4: Length = √(x₁² + y₁² + 2gx₁ + 2fy₁ + c) Step 5: Length = √(3² + 4² + 2(1)(3) + 2((3/2))(4) - 1) = √(9 + 16 + 6 + 12 - 1) = √42

🧠 The One Thing Most Students Get Wrong

Most students get the equation of the tangent to a circle at a given point wrong
They forget to use the correct equation of the tangent, which is xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
This can be avoided by carefully reading the question and using the correct formula

👁️ Ayush's Note

Always check if the given equation represents a circle
Find the centre and radius of the circle
Use the correct equation of the tangent or normal to the circle
Check your calculations carefully to avoid mistakes

🔁 Last 5 Minutes Box

Quickly revise the formulae for the equation of a circle, centre, radius, tangent, and normal
Check if you have used the correct formulae ∈ the solutions
Make sure you have calculated the centre and radius of the circle correctly
Check if you have used the correct equation of the tangent or normal to the circle

📝 Practice MCQs

1. What is the equation of the circle with centre (0, 0) and radius 5?

A) x² + y² = 25

B) x² + y² = 10

C) x² + y² = 5

D) x² + y² = 50

Answer: A) x² + y² = 25

2. What is the centre of the circle x² + y² + 2x + 4y + 4 = 0?

A) (-1, -2)

B) (1, 2)

C) (-2, -4)

D) (2, 4)

Answer: A) (-1, -2)

3. What is the radius of the circle x² + y² + 2x + 3y - 1 = 0?

A) √(1² + (3/2)² + 1) = √(1 + 9/4 + 1) = √(4 + 9)/4 = √13/2

B) √(1² + (3/2)² - 1) = √(1 + 9/4 - 1) = √9/4 = 3/2

C) √(1² + (3/2)² + 1) = √(1 + 9/4 + 1) = √(4 + 9)/4 = √13/2

D) √(1² + (3/2)² - 1) = √(1 + 9/4 - 1) = √9/4 = 3/2

Answer: B) √(1² + (3/2)² - 1) = √(1 + 9/4 - 1) = √9/4 = 3/2

4. What is the equation of the tangent to the circle x² + y² = 25 at point (3, 4)?

A) 3x + 4y = 25

B) 3x + 4y = 50

C) 3x + 4y = 5

D) 3x + 4y = 10

Answer: A) 3x + 4y = 25

5. What is the length of the tangent from point (1, 1) to the circle x² + y² + 2x + 2y + 1 = 0?

A) √(1² + 1² + 2(1) + 2(1) + 1) = √(1 + 1 + 2 + 2 + 1) = √7

B) √(1² + 1² + 2(1) + 2(1) - 1) = √(1 + 1 + 2 + 2 - 1) = √5

C) √(1² + 1² + 2(1) + 2(1) + 1) = √(1 + 1 + 2 + 2 + 1) = √7

D) √(1² + 1² + 2(1) + 2(1) - 1) = √(1 + 1 + 2 + 2 - 1) = √5

Answer: B) √(1² + 1² + 2(1) + 2(1) - 1) = √(1 + 1 + 2 + 2 - 1) = √5

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📚 Academic References

Content verified against peer-reviewed research:

�Let the People Rap�: Cultural Rhetorics Pedagogy and Practices U... — Journal of Basic Writing (2019) 🔓 — DOI ↗
Frustration and Hope: Examining Students� Emotional Responses to ... — Journal of Basic Writing (2019) — DOI ↗
Editors' Column — Journal of Basic Writing (2019) — DOI ↗

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This post was curated by Jules, Exam Compass Bot, and edited for accuracy by Ayush.

📚 Related Topics

Continue your revision with these related guides:

The equation of a circle with centre (h, k) and radius r is (x - h)² + (y - k)² = r²
The equation of a circle with centre (0, 0) and radius r is x² + y² = r²
The general equation of a circle is x² + y² + 2gx + 2fy + c = 0
The centre of the circle x² + y² + 2gx + 2fy + c = 0 is (-g, -f)
The radius of the circle x² + y² + 2gx + 2fy + c = 0 is √(g² + f² - c)
The equation of the tangent to a circle at point (x₁, y₁) is xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
The length of the tangent from point (x₁, y₁) to the circle x² + y² + 2gx + 2fy + c = 0 is √(x₁² + y₁² + 2gx₁ + 2fy₁ + c)
The equation of the normal to a circle at point (x₁, y₁) is y - y₁ = -(x₁/(y₁)) × (x - x₁)