(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.

Question

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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.
**(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.

(4) (i) Find the 2 points and where the
line intersects the circe

(ii) Investigate if is a dianeter
of this circle.

Upstudy AI Solution

Answer

Solution

Extra Insights

Step 1: Substitute

Step 2: Expand and Simplify

Investigation of Diameter

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(4) (i) Find the 2 points and where the line intersects the circe (ii) Investigate if is a dianeter of this circle.