Concept Explanation: Finding the Equation of a Circle

We are given an ellipse equation: $\frac{x^2}{16}+\frac{y^2}{9}=1$. We need to find the equation of a circle that passes through the ellipse's foci and has its center at (0, 3).

Step 1: Find the Foci of the Ellipse

The standard form of an ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a>b$.

Comparing with the given equation: $a^2=16$ and $b^2=9$. Therefore, $a=4$ and $b=3$.

The distance from the center to a focus, $c$, is given by $c=\sqrt{a^2-b^2}$.

Substituting the values: $c=\sqrt{16-9}=\sqrt{7}$.

Since the major axis is along the x-axis, the coordinates of the foci are $(\pm c,0)$ = $(\pm \sqrt{7},0)$.

So, the two foci are: F1 = $(\sqrt{7},0)$ and F2 = $(-\sqrt{7},0)$.

Step 2: General Equation of the Circle

The circle has its center at (0, 3). The general equation of a circle with center (h, k) and radius r is:

$x^2+y^2+2gx+2fy+c=0$.

For center (0, 3), we have $h=0=-g$ and $k=3=-f$. This gives $g=0$ and $f=-3$.

Substituting these values, the equation becomes: $x^2+y^2+2\left(0\right)x+2(-3)y+c=0$.

Simplifying: $x^2+y^2-6y+c=0$.

Our goal is to find the value of the constant $c$.

Step 3: Use the Condition that the Circle Passes Through the Foci

Since the circle passes through both foci, the coordinates of each focus must satisfy the circle's equation. We can use either focus to find $c$.

Let's use the focus F1 = $(\sqrt{7},0)$.

Substitute x = $\sqrt{7}$ and y = 0 into the circle's equation:

${\left(}^{\sqrt{7}}+{\left(}^0-6\left(0\right)+c=0$

This simplifies to: $7+0-0+c=0$

Therefore, $c=-7$.

Step 4: Write the Final Equation

Substitute $c=-7$ back into the general circle equation:

$x^2+y^2-6y+(-7)=0$

Which gives the final equation: $x^2+y^2-6y-7=0$

Final Answer

The equation of the circle is: x² + y² – 6y – 7 = 0

Related Topics & Formulae

Ellipse

Standard Equation (Horizontal Major Axis): $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where a > b.

Foci: Located at $(\pm c,0)$, where $c=\sqrt{a^2-b^2}$.

Circle

Standard Form: $x^2+y^2+2gx+2fy+c=0$

Center: