Questions: On a single set of axes, graph (x^2+y^2=25) and (x^2+10 y^2=169). Solve these equations simultaneously.

Solve the equations simultaneously

Convert the first equation to standard form

The equation \(x^2 + y^2 = 25\) is already in standard form of a circle with center at the origin and radius 5.

Convert the second equation to standard form

The equation \(x^2 + 10y^2 = 169\) is an ellipse. Divide the entire equation by 169 to get \(\frac{x^2}{169} + \frac{y^2}{16.9} = 1\).

Find the intersection points

To find the intersection points, substitute \(x^2 = 25 - y^2\) from the first equation into the second equation: \[ 25 - y^2 + 10y^2 = 169 \] \[ 9y^2 = 144 \] \[ y^2 = 16 \] \[ y = \pm 4 \] Substitute \(y = 4\) and \(y = -4\) back into \(x^2 = 25 - y^2\): For \(y = 4\), \(x^2 = 25 - 16 = 9\), so \(x = \pm 3\). For \(y = -4\), \(x^2 = 25 - 16 = 9\), so \(x = \pm 3\). The intersection points are \((3, 4)\), \((-3, 4)\), \((3, -4)\), and \((-3, -4)\).

\(\boxed{(3, 4), (-3, 4), (3, -4), (-3, -4)}\)

\(\boxed{(3, 4), (-3, 4), (3, -4), (-3, -4)}\)

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