Question
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(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.

Ask by Wheeler Parry. in Ireland
Jan 19,2025

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Answer

**(i)** The points where the line intersects the circle are \( p = (11, 12) \) and \( q = (-13, 4) \). **(ii)** The line segment \( pq \) is a diameter of the circle.

Solution

Let's solve the problem step by step. ### Given: 1. **Line:** \( x - 3y + 25 = 0 \) 2. **Circle:** \( x^2 + y^2 + 2x - 16y - 95 = 0 \) ### Part (i): Finding Points \( p \) and \( q \) of Intersection To find the points where the line intersects the circle, we'll solve the system of equations simultaneously. 1. **Express \( x \) from the line equation:** \[ x - 3y + 25 = 0 \quad \Rightarrow \quad x = 3y - 25 \] 2. **Substitute \( x = 3y - 25 \) into the circle equation:** \[ (3y - 25)^2 + y^2 + 2(3y - 25) - 16y - 95 = 0 \] 3. **Expand and simplify:** \[ 9y^2 - 150y + 625 + y^2 + 6y - 50 - 16y - 95 = 0 \] \[ (9y^2 + y^2) + (-150y + 6y - 16y) + (625 - 50 - 95) = 0 \] \[ 10y^2 - 160y + 480 = 0 \] 4. **Simplify the quadratic equation:** \[ y^2 - 16y + 48 = 0 \quad \text{(Divide by 10)} \] 5. **Solve for \( y \) using the quadratic formula:** \[ y = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 48}}{2 \cdot 1} = \frac{16 \pm \sqrt{256 - 192}}{2} = \frac{16 \pm 8}{2} \] \[ y = \frac{24}{2} = 12 \quad \text{or} \quad y = \frac{8}{2} = 4 \] 6. **Find corresponding \( x \) values:** - For \( y = 12 \): \[ x = 3(12) - 25 = 36 - 25 = 11 \] So, \( p = (11, 12) \). - For \( y = 4 \): \[ x = 3(4) - 25 = 12 - 25 = -13 \] So, \( q = (-13, 4) \). **Therefore, the points of intersection are:** \[ p = (11, 12) \quad \text{and} \quad q = (-13, 4) \] ### Part (ii): Investigating if \( pq \) is a Diameter of the Circle To determine whether the line segment \( pq \) is a diameter of the circle, we need to verify two conditions: 1. **The center of the circle lies on the line \( pq \).** 2. **The segment \( pq \) passes through the center and both endpoints are on the circle.** 1. **Find the center and radius of the circle:** The general equation of a circle is: \[ x^2 + y^2 + 2gx + 2fy + c = 0 \] Comparing with the given equation: \[ x^2 + y^2 + 2x - 16y - 95 = 0 \] We have: \[ 2g = 2 \quad \Rightarrow \quad g = 1 \] \[ 2f = -16 \quad \Rightarrow \quad f = -8 \] The **center** \((h, k)\) of the circle is: \[ h = -g = -1, \quad k = -f = 8 \] So, the center is at \( (-1, 8) \). 2. **Check if the center lies on the line \( pq \):** The line \( pq \) is the same as the given line \( x - 3y + 25 = 0 \). Substitute the center \((-1, 8)\) into the line equation: \[ -1 - 3(8) + 25 = -1 - 24 + 25 = 0 \] Since the equation holds true, the center lies on the line \( pq \). 3. **Check the midpoint of \( pq \):** The midpoint should coincide with the center if \( pq \) is a diameter. Midpoint \( M \) of \( p = (11, 12) \) and \( q = (-13, 4) \): \[ M = \left( \frac{11 + (-13)}{2}, \frac{12 + 4}{2} \right) = \left( \frac{-2}{2}, \frac{16}{2} \right) = (-1, 8) \] This is the center of the circle. **Conclusion:** Since the midpoint of \( pq \) is the center of the circle and both points \( p \) and \( q \) lie on the circle, the line segment \( pq \) is indeed a **diameter** of the circle. ### Final Answer: **(i)** The points of intersection are \( p = (11,\,12) \) and \( q = (-13,\,4) \). **(ii)** The line segment \( pq \) is a diameter of the given 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!

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