Question

Let \( p(x, y) \) be a variable point on the circle \( x^2 + y^2 - 6x - 8y + 21 = 0 \). Then the maximum possible distance from the vertex of \( y^2 + 6y + x + 13 = 0 \) is:

• A. \( 7 + 2\sqrt{2} \)
• B. \( 2 + 7\sqrt{2} \)
• C. \( 4 + 7\sqrt{2} \)
• D. \( 3 + 2\sqrt{2} \)

Hint:

To find the vertex of a parabola quickly, differentiate the squared variable part. Here, \( \frac{d}{dy}(y^2 + 6y) = 2y + 6 = 0 \implies y = -3 \). Then substitute back to find \( x \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To find the maximum distance between a point on a circle and an external point (the vertex), we find the distance between the external point and the centre of the circle, then add the radius.

Step 2: Key Formula or Approach:
1. Circle \( S \): Centre \( C \), Radius \( r \). 2. Vertex \( V \) of the parabola. 3. Max Distance = \( \text{dist}(V, C) + r \).

Step 3: Detailed Explanation:
1. For circle \( x^2 + y^2 - 6x - 8y + 21 = 0 \): Centre \( C = (3, 4) \). Radius \( r = \sqrt{3^2 + 4^2 - 21} = \sqrt{25 - 21} = 2 \). 2. For parabola \( y^2 + 6y + x + 13 = 0 \): Complete the square for \( y \): \( (y+3)^2 - 9 + x + 13 = 0 \implies (y+3)^2 = -(x + 4) \). Vertex \( V = (-4, -3) \). 3. Distance between \( V(-4, -3) \) and \( C(3, 4) \): \[ d = \sqrt{(3 - (-4))^2 + (4 - (-3))^2} = \sqrt{7^2 + 7^2} = 7\sqrt{2} \] 4. Maximum distance = \( d + r = 7\sqrt{2} + 2 \).

Step 4: Final Answer:
The maximum distance is \( 2 + 7\sqrt{2} \).