Find the equation of a circle with centre (-a, -b) and radius √(a2

1.	Find the equation of a circle with centre (-a, -b) and radius √(a2 – b2).
Answer» The general form of the equation of a circle is: (x – h)² + (y – k)² = r² …..(1) Where, r is the radius of the circle and (h, k) is the centre of the circle. Here, r = √(a² – b²), h = -a and k = -b Equation (1)⇒ (x + a )² + (y + b)² = (√(a² – b²))² or (x + a )² + (y + b)²= a²– b² or x² + y² + 2a x + 2ay + a² + b² = a²– b² or x² + y² + 2a x + 2ay + 2 b² = 0 Which is the required equation.

Find the equation of a circle with centre (-a, -b) and radius √(a2 – b2).

Answer»

The general form of the equation of a circle is:

(x – h)² + (y – k)² = r² …..(1)

Where, r is the radius of the circle and (h, k) is the centre of the circle.

Here, r = √(a² – b²), h = -a and k = -b

Equation (1)⇒

(x + a )² + (y + b)² = (√(a² – b²))²

or (x + a )² + (y + b)²= a²– b²

or x² + y² + 2a x + 2ay + a² + b² = a²– b²

or x² + y² + 2a x + 2ay + 2 b² = 0

Which is the required equation.

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Find the equation of a circle with centre (-a, -b) and radius √(a2 – b2).

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