Let's solve each problem step by step and explain the reasoning.
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Problem 1:
Write the Standard Form of the equation for a circle centered at point (4, 6) and has a radius of 10.
#### 🔹 Standard Form of a Circle:
The standard form of a circle with center $(h, k)$ and radius $r$ is:
$$
(x - h)^2 + (y - k)^2 = r^2
$$
Given:
- Center: $(h, k) = (4, 6)$
- Radius: $r = 10$, so $r^2 = 100$
Plug in the values:
$$
(x - 4)^2 + (y - 6)^2 = 100
$$
✔ Correct answer:
(x - 4)^2 + (y - 6)^2 = 100
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Problem 2:
Write the Standard Form of the equation for a circle centered at point (-2, -8) and has a radius of 4.
Using the same formula:
$$
(x - h)^2 + (y - k)^2 = r^2
$$
Given:
- Center: $(-2, -8)$ → $h = -2$, $k = -8$
- Radius: $r = 4$, so $r^2 = 16$
Now plug in:
$$
(x - (-2))^2 + (y - (-8))^2 = 16
\Rightarrow (x + 2)^2 + (y + 8)^2 = 16
$$
✔ Correct answer:
(x + 2)^2 + (y + 8)^2 = 16
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Problem 3:
Convert the following Standard Form equation to General Form:
$$
(x - 7)^2 + (y + 3)^2 = 49
$$
We need to expand this and write it in
General Form, which is:
$$
x^2 + y^2 + Dx + Ey + F = 0
$$
#### Step 1: Expand both squared terms
$$
(x - 7)^2 = x^2 - 14x + 49
$$
$$
(y + 3)^2 = y^2 + 6y + 9
$$
Add them:
$$
x^2 - 14x + 49 + y^2 + 6y + 9 = 49
$$
Combine like terms:
$$
x^2 + y^2 - 14x + 6y + 58 = 49
$$
Subtract 49 from both sides:
$$
x^2 + y^2 - 14x + 6y + 9 = 0
$$
✔ Correct answer:
x² + y² - 14x + 6y + 9 = 0
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✔ Final Answers:
1.
(x - 4)^2 + (y - 6)^2 = 100
2.
(x + 2)^2 + (y + 8)^2 = 16
3.
x² + y² - 14x + 6y + 9 = 0
Let me know if you'd like a visual explanation or help with more problems!
Parent Tip: Review the logic above to help your child master the concept of circle equation worksheet.
