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:
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:
Updated On: Apr 12, 2026
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
We are asked to find the maximum distance between a point P, which can be anywhere on a given circle, and a fixed point, which is the vertex of a given parabola.
Step 2: Key Formula or Approach:
1. Find the standard form of the circle's equation to determine its center and radius. The standard form is $(x-h)^2 + (y-k)^2 = r^2$.
2. Find the standard form of the parabola's equation to determine its vertex. The standard form for a vertical parabola is $(x-h)^2 = 4p(y-k)$ or $(x-h)^2 = -4p(y-k)$.
3. The maximum distance from a fixed external point to a circle is the sum of the distance from the point to the circle's center and the circle's radius. Max Distance = $d(Point, Center) + r$.
Step 3: Detailed Explanation:
Analyze the Circle:
The equation of the circle is $x^2 + y^2 - 6x - 8y + 21 = 0$.
We complete the square for the x and y terms:
$(x^2 - 6x) + (y^2 - 8y) + 21 = 0$
$(x^2 - 6x + 9) - 9 + (y^2 - 8y + 16) - 16 + 21 = 0$
$(x-3)^2 + (y-4)^2 = 9 + 16 - 21$
$(x-3)^2 + (y-4)^2 = 4 = 2^2$
So, the center of the circle is $C = (3, 4)$ and the radius is $r = 2$.
Analyze the Parabola:
The equation of the parabola is $x^2 + 6x + y + 13 = 0$.
We complete the square for the x terms to find the vertex:
$(x^2 + 6x) + y + 13 = 0$
$(x^2 + 6x + 9) - 9 + y + 13 = 0$
$(x+3)^2 + y + 4 = 0$
$(x+3)^2 = -(y+4)$
This is a downward-opening parabola with its vertex at $V = (-3, -4)$.
Find the Maximum Distance:
We need to find the maximum distance between any point P on the circle and the fixed point V(-3, -4).
The maximum distance from an external point to a circle lies on the line passing through the point and the center of the circle.
First, calculate the distance between the vertex V and the center of the circle C.
$d(V, C) = \sqrt{(3 - (-3))^2 + (4 - (-4))^2}$
$d(V, C) = \sqrt{(6)^2 + (8)^2}$
$d(V, C) = \sqrt{36 + 64} = \sqrt{100} = 10$.
The maximum distance from V to a point P on the circle is the distance from V to the center C plus the radius r.
Max Distance = $d(V, C) + r$
Max Distance = $10 + 2 = 12$.
This occurs at the point P on the circle that lies on the line segment extending from V through C.
Step 4: Final Answer:
The maximum distance of P from the vertex of the parabola is 12.
We are asked to find the maximum distance between a point P, which can be anywhere on a given circle, and a fixed point, which is the vertex of a given parabola.
Step 2: Key Formula or Approach:
1. Find the standard form of the circle's equation to determine its center and radius. The standard form is $(x-h)^2 + (y-k)^2 = r^2$.
2. Find the standard form of the parabola's equation to determine its vertex. The standard form for a vertical parabola is $(x-h)^2 = 4p(y-k)$ or $(x-h)^2 = -4p(y-k)$.
3. The maximum distance from a fixed external point to a circle is the sum of the distance from the point to the circle's center and the circle's radius. Max Distance = $d(Point, Center) + r$.
Step 3: Detailed Explanation:
Analyze the Circle:
The equation of the circle is $x^2 + y^2 - 6x - 8y + 21 = 0$.
We complete the square for the x and y terms:
$(x^2 - 6x) + (y^2 - 8y) + 21 = 0$
$(x^2 - 6x + 9) - 9 + (y^2 - 8y + 16) - 16 + 21 = 0$
$(x-3)^2 + (y-4)^2 = 9 + 16 - 21$
$(x-3)^2 + (y-4)^2 = 4 = 2^2$
So, the center of the circle is $C = (3, 4)$ and the radius is $r = 2$.
Analyze the Parabola:
The equation of the parabola is $x^2 + 6x + y + 13 = 0$.
We complete the square for the x terms to find the vertex:
$(x^2 + 6x) + y + 13 = 0$
$(x^2 + 6x + 9) - 9 + y + 13 = 0$
$(x+3)^2 + y + 4 = 0$
$(x+3)^2 = -(y+4)$
This is a downward-opening parabola with its vertex at $V = (-3, -4)$.
Find the Maximum Distance:
We need to find the maximum distance between any point P on the circle and the fixed point V(-3, -4).
The maximum distance from an external point to a circle lies on the line passing through the point and the center of the circle.
First, calculate the distance between the vertex V and the center of the circle C.
$d(V, C) = \sqrt{(3 - (-3))^2 + (4 - (-4))^2}$
$d(V, C) = \sqrt{(6)^2 + (8)^2}$
$d(V, C) = \sqrt{36 + 64} = \sqrt{100} = 10$.
The maximum distance from V to a point P on the circle is the distance from V to the center C plus the radius r.
Max Distance = $d(V, C) + r$
Max Distance = $10 + 2 = 12$.
This occurs at the point P on the circle that lies on the line segment extending from V through C.
Step 4: Final Answer:
The maximum distance of P from the vertex of the parabola is 12.
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