The equation of the circle drawn with the two foci of x^2 a^2 + y… - Vidyadip

MCQ

The equation of the circle drawn with the two foci of $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$ end-point of a diameter is

A
$\text{x}^2+\text{y}^2=\text{a}^2+\text{b}^2$
B
$\text{x}^2+\text{y}^2=\text{a}^2$
C
$\text{x}^2+\text{y}^2=2\text{a}^2$
✓
$\text{x}^2+\text{y}^2=\text{a}^2-\text{b}^2$

✓

Answer

Correct option: D.

$\text{x}^2+\text{y}^2=\text{a}^2-\text{b}^2$

We have $\text{r}=\text{ae}$
Let the equation of the circle be $\text{x}^2+\text{y}^2=\text{r}^2.$
Now, $\text{x}^2+\text{y}^2=\text{a}^2\text{e}^2$ $(\because\text{r}=\text{ae})$
$\Rightarrow\text{x}^2+\text{y}^2=\text{a}^2\Big(1-\frac{\text{b}^2}{\text{a}^2}\Big)$ $\bigg(\because\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}\bigg)$
$\Rightarrow\text{x}^2+\text{y}^2=\text{a}^2-\text{b}^2$
$\therefore\ $The required equation of the circle $\text{x}^2+\text{y}^2=\text{a}^2-\text{b}^2.$

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