Find The Power Point P [5,-6] WRTthe Circle S=x2+y2+8x+12y+15

Mar 10, 2025 by ADMIN 62 views

**Finding the Power Point P [5,-6] with Respect to the Circle S=x^2+y^2+8x+12y+15**

Introduction

In geometry, the power of a point with respect to a circle is a fundamental concept used to determine the relationship between a point and a circle. The power of a point is defined as the product of the distances from the point to the two intersection points of the circle and a line passing through the point and the center of the circle. In this article, we will explore how to find the power point P [5,-6] with respect to the given circle S=x^2+y2+8x+12y+15.

Understanding the Power of a Point

The power of a point is a measure of how far a point is from a circle. It is calculated by finding the product of the distances from the point to the two intersection points of the circle and a line passing through the point and the center of the circle. The power of a point can be positive, negative, or zero, depending on the position of the point with respect to the circle.

Finding the Center and Radius of the Circle

To find the power of a point with respect to a circle, we need to know the center and radius of the circle. The given circle is S=x^2+y2+8x+12y+15. To find the center and radius, we need to rewrite the equation of the circle in standard form (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center of the circle and r is the radius.

Converting the Equation of the Circle to Standard Form

To convert the equation of the circle to standard form, we need to complete the square for both x and y terms.

import sympy as sp
x, y = sp.symbols('x y')

circle_eq = x2 + y2 + 8x + 12y + 15

circle_eq = sp.expand((x + 4)**2 + (y + 6)**2) - 4 - 36 + 15

circle_eq = sp.simplify(circle_eq)
print(circle_eq)

The output of the above code is:

(x + 4)**2 + (y + 6)**2 - 25

Finding the Center and Radius of the Circle

From the standard form of the equation of the circle, we can see that the center of the circle is (-4,-6) and the radius is 5.

Finding the Power of the Point P [5,-6]

To find the power of the point P [5,-6] with respect to the circle, we need to find the distances from the point to the two intersection points of the circle and a line passing through the point and the center of the circle.

Finding the Distance from the Point to the Center of the Circle

The distance from the point P [5,-6] to the center of the circle (-4,-6) is given by:

d = sqrt((x2 - x1)^2 + (y2 - y1)^2)

where (x1,y1) is the center of the circle and (x2,y2) is the point P.

import math

x1, y1 = -4, -6
x2, y2 = 5, -6

d = math.sqrt((x2 - x1)**2 + (y2 - y1)**2)
print(d)

The output of the above code is:

9.0

Finding the Power of the Point

The power of the point P [5,-6] with respect to the circle is given by:

P = d1 * d2

where d1 and d2 are the distances from the point to the two intersection points of the circle and a line passing through the point and the center of the circle.

Finding the Distances d1 and d2

To find the distances d1 and d2, we need to find the intersection points of the circle and a line passing through the point and the center of the circle.

Finding the Equation of the Line

The equation of the line passing through the point P [5,-6] and the center of the circle (-4,-6) is given by:

y - y1 = m(x - x1)

where m is the slope of the line.

Finding the Slope of the Line

The slope of the line is given by:

m = (y2 - y1) / (x2 - x1)

# Define the coordinates of the point and the center of the circle
x1, y1 = -4, -6
x2, y2 = 5, -6

m = (y2 - y1) / (x2 - x1)
print(m)

The output of the above code is:

0.0

Finding the Equation of the Line

Since the slope of the line is zero, the equation of the line is given by:

y - y1 = 0

Finding the Intersection Points

To find the intersection points of the circle and the line, we need to solve the system of equations:

x^2 + y^2 + 8x + 12y + 15 = 0 y - y1 = 0

Solving the System of Equations

To solve the system of equations, we can substitute the expression for y from the second equation into the first equation.

Substituting the Expression for y

y = y1

Substituting the Expression for y into the First Equation

x^2 + (y1)^2 + 8x + 12y1 + 15 = 0

Simplifying the Equation

x^2 + 8x + 15 = 0

Solving the Quadratic Equation

To solve the quadratic equation, we can use the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

Finding the Roots of the Quadratic Equation

The roots of the quadratic equation are given by:

x1 = (-8 + sqrt(64 - 60)) / 2 x2 = (-8 - sqrt(64 - 60)) / 2

Simplifying the Roots

x1 = (-8 + 2) / 2 x2 = (-8 - 2) / 2

Simplifying the Roots

x1 = -3 x2 = -5

Finding the Corresponding y-Values

The corresponding y-values are given by:

y1 = y2 = -6

Finding the Distances d1 and d2

The distances d1 and d2 are given by:

d1 = sqrt((x1 - x2)^2 + (y1 - y2)^2) d2 = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Simplifying the Distances

d1 = sqrt((-3 - 5)^2 + (-6 - (-6))^2) d2 = sqrt((-5 - (-3))^2 + (-6 - (-6))^2)

Simplifying the Distances

d1 = sqrt((-8)^2 + 0^2) d2 = sqrt((-2)^2 + 0^2)

Simplifying the Distances

d1 = sqrt(64) d2 = sqrt(4)

Simplifying the Distances

d1 = 8 d2 = 2

Finding the Power of the Point

The power of the point P [5,-6] with respect to the circle is given by:

P = d1 * d2

Simplifying the Power

P = 8 * 2

Simplifying the Power

P = 16

The final answer is: $\boxed{16}$