91Ó°ÊÓ

We have the two points (2, 3) and (4, 7) on the line. First, we find the slope (m) of the line using the formula: \(m = \frac{y_2 - y_1}{x_2 - x_1}\) Here, (x1, y1) = (2, 3) and (x2, y2) = (4, 7). So, \(m = \frac{7 - 3}{4 - 2} = \frac{4}{2} = 2\). Now that we have the slope, we can find the equation of the line using the point-slope form, which is given as: \(y - y_1 = m(x - x_1)\) Using the point (2, 3) and the slope m = 2, we get: \(y - 3 = 2(x - 2)\) Now, let's rewrite this equation in the slope-intercept form: \(y = 2x - 1\)

The circle is centered at (3, -3) with a radius of \(\sqrt{15}\). The equation of a circle with center (h, k) and radius r is given as: \((x - h)^2 + (y - k)^2 = r^2\) Plugging in the values for h, k, and r, we get: \((x - 3)^2 + (y + 3)^2 = (\sqrt{15})^2\) Which simplifies to: \((x - 3)^2 + (y + 3)^2 = 15\)

We have the following system of equations: 1. \(y = 2x - 1\) 2. \((x - 3)^2 + (y + 3)^2 = 15\) To solve this system, let's substitute the y value from equation 1 into equation 2: \((x - 3)^2 + ((2x - 1) + 3)^2 = 15\) Simplifying the equation, we get: \((x - 3)^2 + (2x + 2)^2 = 15\) Expanding the equation, we get: \(x^2 - 6x + 9 + 4x^2 + 8x + 4 = 15\) Combining like terms, we get: \(5x^2 + 2x - 2 = 0\) Now, we solve this quadratic equation for x using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Plugging in the values for a, b, and c, we get: \(x = \frac{-2 \pm \sqrt{(-2)^2 - 4(5)(-2)}}{2(5)} = \frac{-2 \pm \sqrt{44}}{10}\) Now we have two x values, let's find their corresponding y values using equation 1: 1. \(y_1 = 2(\frac{-2 + \sqrt{44}}{10}) - 1\) 2. \(y_2 = 2(\frac{-2 - \sqrt{44}}{10}) - 1\) Simplifying the y values, we get: 1. \(y_1 = \frac{\sqrt{44} - 4}{5}\) 2. \(y_2 = \frac{-4 - \sqrt{44}}{5}\) Thus, the intersection points are: \((\frac{-2 + \sqrt{44}}{10}, \frac{\sqrt{44} - 4}{5})\) and \((\frac{-2 - \sqrt{44}}{10}, \frac{-4 - \sqrt{44}}{5})\).