Question 49 of 70intermediate🔧 ApplyShort Answer1 mark

In each of the following Exercises 1 to 5, find the equation of the circle with centre (-a, -b) and radius a2−b2\sqrt{a² - b²}

Correct Answer

(x + a)² + (y + b)² = a² - b²

Exercise: EXERCISE 10.1 | Q: 5 | (Chapter: 6)
For More Understanding

Explanation

This question requires applying the standard equation of a circle formula. The standard form is (x - h)² + (y - k)² = r², where (h, k) is the centre and r is the radius. Given centre (-a, -b) and radius a2−b2\sqrt{a² - b²}, we substitute h = -a, k = -b, and r² = a² - b² into the formula.

Solution Steps

  1. Step 1: Recall the standard equation of a circle: (x - h)² + (y - k)² = r²

  2. Step 2: Identify the given values: Centre (h, k) = (-a, -b) and Radius r = a2−b2\sqrt{a² - b²}

  3. Step 3: Substitute h = -a, k = -b into the equation: (x - (-a))² + (y - (-b))² = r²

  4. Step 4: Simplify: (x + a)² + (y + b)² = r²

  5. Step 5: Calculate r² = (a2−b2\sqrt{a² - b²})² = a² - b²

  6. Step 6: Final equation: (x + a)² + (y + b)² = a² - b²

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