(4) (i) Find the 2 points \( p \) and \( q \) where the line \( x-3 y+25=0 \) intersects the circe \( x^{2}+y^{2}+2 x-16 y-95=0 \) (ii) Investigate if \( p q \) is a dianeter of this circle.
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To start, let’s express the line equation in slope-intercept form: 1. Rearranging \( x - 3y + 25 = 0 \) gives us \( y = \frac{1}{3}x + \frac{25}{3} \). Now, we need to substitute this expression for \( y \) into the circle's equation: 2. The given circle equation \( x^2 + y^2 + 2x - 16y - 95 = 0 \) can be rewritten by plugging in \( y = \frac{1}{3}x + \frac{25}{3} \). This will involve simplifying the equation to find the intersection points. ### Step 1: Substitute \( y \) Replace \( y \): \[ x^2 + \left( \frac{1}{3}x + \frac{25}{3} \right)^2 + 2x - 16 \left( \frac{1}{3}x + \frac{25}{3} \right) - 95 = 0 \] ### Step 2: Expand and Simplify Expanding \( \left( \frac{1}{3}x + \frac{25}{3} \right)^2 \): \[ \frac{1}{9}x^2 + \frac{50}{9}x + \frac{625}{9} \] Plug back into the circle equation and combine like terms to form a quadratic in \( x \). Solving this quadratic will yield the intersection points \( p \) and \( q \). ### Investigation of Diameter To check if the segment \( pq \) is a diameter, we need to determine the center of the circle and then check if \( p \) and \( q \) are opposite points on the circle with respect to this center. The center of the circle can be found by completing the square based on its equation, leading to the identification of its center coordinates. If \( p \) and \( q \) are symmetric about this center, then they indeed form a diameter of the circle. This involves comparing the midpoint of \( pq \) with the center of the circle. If they match, \( pq \) is a diameter!