If the foci of an ellipse `(x^(2))/(9)+(y^(2))/(16)=1` are ends of diameter of a circle then finds the radius of the circle

A circle is drawn with the two foci an ellipse x^(2)/a^(2)+y^(2)/b^(2)=1 at the end of the diameter. What is the equation of the eircle ?

The foci of the hyperbola (x^(2))/(16) -(y^(2))/(9) = 1 is :

Find the foci of the ellipse 25(x+1)^(2)+9(y+2)^(2)=225

The circle described on the line joining the foci of the hyperbola (x^(2))/(16)-(y^(2))/(9) = 1 as a diameter passes through an end of the latus rectum of the parabola y^(2) = 4ax , the length of the latus rectum of the parabola is

The length of the tangent of the ellipse (x^(2))/(25)+(y^(2))/(16)=1 intercepted between auxiliary circle such that the portionof the tangent intercepted between the auxiliary circle subtends equal angles at foci is

Find the foci and eccentricity of ellipse (x^(2))/(16) + (y^(2))/(4) = 1 .

The ellipse (x^(2))/(16)+(y^(2))/(7)=1 has its foci at S and s'. Find the area of the ellipse falling outsidethe circle whose diameter is SS'.