The line `3x -y -17=0` meets the circle `x ^(2) +y ^(2) -8x+ 10 y + 5=0` at the point A and B. P is any point on the circle other than A or B, then the triangle APR is

Options

A. equilateral
B. isosceles
C. right angled
D. obtuse angled

Answer

To solve the problem step by step, we will analyze the given line and circle equations, find the points of intersection, and determine the nature of triangle APR.

### Step 1: Rewrite the Circle's Equation
The equation of the circle is given as:
\[ x^2 + y^2 - 8x + 10y + 5 = 0 \]

We can rewrite it in standard form by completing the square.

1. Group the x and y terms:
\[ (x^2 - 8x) + (y^2 + 10y) + 5 = 0 \]

2. Complete the square for x:
\[ x^2 - 8x = (x - 4)^2 - 16 \]

3. Complete the square for y:
\[ y^2 + 10y = (y + 5)^2 - 25 \]

4. Substitute back into the equation:
\[ (x - 4)^2 - 16 + (y + 5)^2 - 25 + 5 = 0 \]
\[ (x - 4)^2 + (y + 5)^2 - 36 = 0 \]
\[ (x - 4)^2 + (y + 5)^2 = 36 \]

This shows that the circle is centered at \( (4, -5) \) with a radius of \( 6 \).

### Step 2: Find the Points of Intersection (A and B)
The line equation is given as:
\[ 3x - y - 17 = 0 \]
We can express y in terms of x:
\[ y = 3x - 17 \]

Now, substitute this expression for y into the circle's equation:
\[ (x - 4)^2 + (3x - 17 + 5)^2 = 36 \]
\[ (x - 4)^2 + (3x - 12)^2 = 36 \]

Expanding both squares:
1. \( (x - 4)^2 = x^2 - 8x + 16 \)
2. \( (3x - 12)^2 = 9x^2 - 72x + 144 \)

Combine these:
\[ x^2 - 8x + 16 + 9x^2 - 72x + 144 = 36 \]
\[ 10x^2 - 80x + 160 = 36 \]
\[ 10x^2 - 80x + 124 = 0 \]
\[ x^2 - 8x + 12.4 = 0 \]

Now, we can use the quadratic formula to find x:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1, b = -8, c = 12.4 \):
\[ x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 12.4}}{2 \cdot 1} \]
\[ x = \frac{8 \pm \sqrt{64 - 49.6}}{2} \]
\[ x = \frac{8 \pm \sqrt{14.4}}{2} \]

Calculating the square root:
\[ \sqrt{14.4} \approx 3.8 \]
Thus,
\[ x_1 = \frac{8 + 3.8}{2} \approx 5.9, \quad x_2 = \frac{8 - 3.8}{2} \approx 2.1 \]

Now substitute these x-values back into the line equation to find corresponding y-values.

### Step 3: Determine the Nature of Triangle APR
Since the line passes through the center of the circle, points A and B are endpoints of the diameter. According to the property of circles, any angle subtended by a diameter at the circumference is a right angle.

Thus, triangle APR is a right triangle.

### Conclusion
The triangle APR is a right triangle.