To find the radius of the circle that passes through the foci of the given ellipse \(9x^2 + 16y^2 = 144\) and has its center at \((0,3)\), we need to follow these steps:
First, rewrite the equation of the ellipse in its standard form. The given equation is: \(9x^2 + 16y^2 = 144\) Divide through by 144 to get: \(\frac{x^2}{16} + \frac{y^2}{9} = 1\) This is an ellipse equation of the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a^2 = 16\) and \(b^2 = 9\).
Identify the orientation and components of the ellipse. Here, \(a = 4\) and \(b = 3\). Since \(a > b\), it is a horizontal ellipse.
The foci of the ellipse are located at \((\pm c, 0)\) where \(c\) is calculated by: \(c^2 = a^2 - b^2\) Calculate \( c \): \(c^2 = 16 - 9 = 7 \rightarrow c = \sqrt{7}\) Therefore, the foci are located at \((\pm \sqrt{7}, 0)\).
The equation of a circle with center \((h, k)\) and radius \(r\) is given by: \((x-h)^2 + (y-k)^2 = r^2\) For the circle centered at \((0,3)\), its equation becomes: \(x^2 + (y-3)^2 = r^2\)
Since the circle passes through the points \((\sqrt{7}, 0)\) and \((- \sqrt{7}, 0)\), substitute one of the foci: \(\sqrt{7}^2 + (0-3)^2 = r^2\) Simplify: \(r^2 = 7 + 9 = 16 \rightarrow r = \sqrt{16} = 4\)