Question:medium

The equation of a circle whose Centre is $(-3, 2)$ and area is $176$ units is:

Show Hint

Center $(h, k)$ means the equation starts with $x^2 + y^2 - 2hx - 2ky$. For $(-3, 2)$, it's $+6x - 4y$.
  • $x^2 + y^2 + 6x - 4y - 36 = 0$
  • $x^2 + y^2 + 6x - 4y - 43 = 0$
  • $x^2 + y^2 - 6x + 4y - 36 = 0$
  • $x^2 + y^2 - 6x + 4y - 43 = 0$
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
The equation of a circle with centre \( (h, k) \) and radius \( r \) is \( (x-h)^2 + (y-k)^2 = r^2 \). We are given the area, which allows us to calculate \( r^2 \).
Step 2: Key Formula or Approach:
1. Area of circle = \( \pi r^2 \). (Use \( \pi \approx 22/7 \) for calculation).
2. Standard form expansion: \( x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0 \).
Step 3: Detailed Explanation:
Given Area = 176: \[ \pi r^2 = 176 \implies \frac{22}{7} r^2 = 176 \] \[ r^2 = \frac{176 \times 7}{22} = 8 \times 7 = 56 \] Substitute centre \( (h, k) = (-3, 2) \) and \( r^2 = 56 \) into the circle equation: \[ (x - (-3))^2 + (y - 2)^2 = 56 \] \[ (x + 3)^2 + (y - 2)^2 = 56 \] Expand the terms: \[ (x^2 + 6x + 9) + (y^2 - 4y + 4) = 56 \] \[ x^2 + y^2 + 6x - 4y + 13 - 56 = 0 \] \[ x^2 + y^2 + 6x - 4y - 43 = 0 \]
Step 4: Final Answer:
The equation of the circle is \( x^2 + y^2 + 6x - 4y - 43 = 0 \).
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