Question

The equation of the circle drawn with the two foci of x^2/a^2+y^2/b^2=1 as the end points of a diameter is

Answer 1

Given info : the equation of the circle drawn with the foci of x²/a² + y²/b² = 1 as the end points of a diameter is...

Solution : a/c to question, diameter of circle = length of major axis

⇒2r = 2ae

⇒r = ae

Here e is eccentricity of ellipse.

Now but we know e = √(1 - b²/a²)

so, r = a √(1 - b²/a²) = a√(a² - b²)/a = √(a² - b²)

So equation of circle is (x - 0)² + (y - 0)² = (√(a² - b²))² = (a² - b²)

[ Origin of circle is taken (0,0) because equation of ellipse is (x - 0)/a² + (y - 0)/b² = 1 ]

Therefore the equation of circle is x² + y² = a² - b²

Answer 2

Answer:-

Given info :

the equation of the circle drawn with the foci of x²/a² + y²/b² = 1 as the end points of a diameter is...

Solution :

a/c to question, diameter of circle = length of major axis

⇒2r = 2ae

⇒r = ae

Here e is eccentricity of ellipse.

Now but we know e = √(1 - b²/a²)

so,

r = a √(1 - b²/a²) = a√(a² - b²)/a = √(a² - b²)

So equation of circle is (x - 0)² + (y - 0)² = (√(a² - b²))² = (a² - b²)

[ Origin of circle is taken (0,0) because equation of ellipse is (x - 0)/a² + (y - 0)/b² = 1 ]

Therefore the equation of circle is x² + y² = a² - b²

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