Question

Let P be a moving point on the circle $x^2 + y^2-6x-8y + 21 = 0$. Then, the maximum distance of P from the vertex of the parabola $x^2 + 6x + y + 13 = 0$ is equal to:

• A. 8
• B. 10
• C. 12
• D. 9

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the greatest possible distance between a point on a given circle and a fixed point, where the fixed point is the vertex of a given parabola.
Step 2: Key Formula or Approach:
1. Identify Circle's Center and Radius: Convert the general form of the circle's equation to the standard form $(x-h)^2 + (y-k)^2 = r^2$ by completing the square.
2. Identify Parabola's Vertex: Convert the general form of the parabola's equation to the standard vertex form $(x-h)^2 = \pm 4p(y-k)$ by completing the square.
3. Geometric Interpretation: The maximum distance from an external point to a circle is along the line connecting the point and the circle's center. The distance is the sum of the distance between the point and the center, and the circle's radius.
Max Distance = (Distance from external point to center) + (Radius).
Step 3: Detailed Explanation:
Part 1: Analyze the Circle
Given circle equation: $x^2 + y^2 - 6x - 8y + 21 = 0$.
Group terms and complete the square:
$(x^2 - 6x + 9) + (y^2 - 8y + 16) - 9 - 16 + 21 = 0$
$(x-3)^2 + (y-4)^2 = 4 = 2^2$.
The center of the circle is $C = (3, 4)$ and the radius is $r = 2$.
Part 2: Analyze the Parabola
Given parabola equation: $x^2 + 6x + y + 13 = 0$.
Isolate and complete the square for x terms:
$x^2 + 6x + 9 = -y - 13 + 9$
$(x+3)^2 = -y - 4 \implies (x+3)^2 = -(y+4)$.
The vertex of the parabola is $V = (-3, -4)$.
Part 3: Calculate the Maximum Distance
We need the maximum distance from the fixed point $V(-3, -4)$ to any point on the circle centered at $C(3, 4)$ with radius 2.
First, find the distance between the vertex V and the center C using the distance formula:
\[ d(V, C) = \sqrt{(3 - (-3))^2 + (4 - (-4))^2} = \sqrt{6^2 + 8^2} \] \[ d(V, C) = \sqrt{36 + 64} = \sqrt{100} = 10 \] The maximum distance from V to a point on the circle is the distance to the center plus the radius:
\[ \text{Max Distance} = d(V, C) + r = 10 + 2 = 12 \] Step 4: Final Answer:
The maximum distance of P from the vertex of the parabola is 12.