Question

The slope of the straight line joining the centre of the circle \( x^2 + y^2 - 8x + 2y = 0 \) and the vertex of the parabola \( y = x^2 - 4x + 10 \) is:

• A. \( \frac{-5}{2} \)
• B. \( \frac{-7}{2} \)
• C. \( \frac{-3}{2} \)
• D. \( \frac{5}{2} \)
• E. \( \frac{7}{2} \)

Hint:

To find the vertex of \( y = f(x) \) quickly, set \( f'(x) = 0 \). Here, \( y' = 2x - 4 = 0 \Rightarrow x = 2 \).

Answer 1

Answer: (E)

Concept: The center of a circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is \( (-g, -f) \). The vertex of a parabola \( y = ax^2 + bx + c \) occurs at \( x = -b/(2a) \). The slope \( m \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Step 1: Find the centre of the circle.
Equation: \( x^2 + y^2 - 8x + 2y = 0 \) By comparison: \( 2g = -8 \Rightarrow g = -4 \) and \( 2f = 2 \Rightarrow f = 1 \). Centre \( C = (-g, -f) = (4, -1) \).

Step 2: Find the vertex of the parabola.
Equation: \( y = x^2 - 4x + 10 \) The x-coordinate of the vertex is \( x = \frac{-(-4)}{2(1)} = 2 \). Substitute \( x = 2 \) into the equation to find \( y \): \[ y = (2)^2 - 4(2) + 10 = 4 - 8 + 10 = 6 \] Vertex \( V = (2, 6) \).

Step 3: Calculate the slope between (4, -1) and (2, 6).
\[ m = \frac{6 - (-1)}{2 - 4} = \frac{7}{-2} = -\frac{7}{2} \]