Math Problem Statement

Find the equations and solve the given problems: 1. Solve x^2 + y^2 - 7x + 5y - 14 = 0, passing through (5, 3). 2. Check if the point (-7, -2) lies on the circle (x+1)^2 + (y-2)^2 = 26. 3. Find the tangential distance from a point to the circle (x+1)^2 + (y-2)^2 = 26. 4. Rewrite and solve for the equation of a parabola: x^2 + y^2 - 4x - 12 = 0. 5. Find the equation of the line y = z passing through (0, 4) and (3, -2).

Solution

Let me extract and solve each part of the question from the uploaded image. Here's what the image includes:

Problems

  1. Solve for the equation x2+y27x+5y14=0x^2 + y^2 - 7x + 5y - 14 = 0, passing through (5,3)(5, 3).
  2. Confirm if (7,2)(-7, -2) lies on the circle (x+1)2+(y2)2=26(x+1)^2 + (y-2)^2 = 26.
  3. Find the tangential distance from a point to the given circle.
  4. Solve for the equation of a parabola: x2+y24x12=0x^2 + y^2 - 4x - 12 = 0.
  5. Find the equation of the line y=zy = z, passing through points (0,4)(0, 4) and (3,2)(3, -2).

Step-by-step Solutions:

1. Solve x2+y27x+5y14=0x^2 + y^2 - 7x + 5y - 14 = 0, given point (5,3)(5, 3):

Substitute (x,y)=(5,3)(x, y) = (5, 3) into the equation to check if the point satisfies it: x2+y27x+5y14=52+327(5)+5(3)14x^2 + y^2 - 7x + 5y - 14 = 5^2 + 3^2 - 7(5) + 5(3) - 14 =25+935+1514=0= 25 + 9 - 35 + 15 - 14 = 0 Since the result is 0, (5,3)(5, 3) lies on the curve.


2. Check if (7,2)(-7, -2) lies on the circle (x+1)2+(y2)2=26(x+1)^2 + (y-2)^2 = 26:

Substitute (x,y)=(7,2)(x, y) = (-7, -2) into the circle equation: (x+1)2+(y2)2=(7+1)2+(22)2=(6)2+(4)2=36+16=52(x+1)^2 + (y-2)^2 = (-7+1)^2 + (-2-2)^2 = (-6)^2 + (-4)^2 = 36 + 16 = 52 The result is 52, not 26. Hence, (7,2)(-7, -2) does not lie on the circle.


3. Find tangential distance from a point to the circle:

Let the circle be (x+1)2+(y2)2=26(x+1)^2 + (y-2)^2 = 26, with center (1,2)(-1, 2) and radius 26\sqrt{26}. The tangential distance from a point (x1,y1)(x_1, y_1) to the circle is: Distance=(x1h)2+(y1k)2r\text{Distance} = \left| \sqrt{(x_1 - h)^2 + (y_1 - k)^2} - r \right| Where (h,k)(h, k) is the center of the circle. Let me know the point from which the tangential distance is needed.


4. Solve for the parabola x2+y24x12=0x^2 + y^2 - 4x - 12 = 0:

Rewrite in standard form: x24x+y2=12x^2 - 4x + y^2 = 12 Complete the square for xx: (x24x+4)+y2=12+4    (x2)2+y2=16(x^2 - 4x + 4) + y^2 = 12 + 4 \quad \implies \quad (x-2)^2 + y^2 = 16 This is a circle with center (2,0)(2, 0) and radius 44, not a parabola.


5. Equation of the line passing through (0,4)(0, 4) and (3,2)(3, -2):

The slope mm of the line is: m=2430=63=2m = \frac{-2 - 4}{3 - 0} = \frac{-6}{3} = -2 Using the point-slope form of the equation yy1=m(xx1)y - y_1 = m(x - x_1), with (x1,y1)=(0,4)(x_1, y_1) = (0, 4): y4=2(x0)    y=2x+4y - 4 = -2(x - 0) \quad \implies \quad y = -2x + 4 Thus, the line's equation is y=2x+4y = -2x + 4.


Let me know if you'd like any clarifications or further details on these solutions!

Follow-up Questions:

  1. How do you find the center and radius of a circle from its general equation?
  2. What is the formula for finding the tangential distance from a point to a circle?
  3. How do you confirm if a point lies on a given curve or circle?
  4. What is the method for completing the square for conic sections?
  5. How do you derive the equation of a line from two points?

Tip:

Always verify your solutions by substituting back into the original equations to ensure correctness.

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Math Problem Analysis

Mathematical Concepts

Analytic Geometry
Equations of Circles
Distance Formula
Equation of a Line
Equation of a Parabola

Formulas

General equation of a circle: (x - h)^2 + (y - k)^2 = r^2
Distance formula: sqrt((x2 - x1)^2 + (y2 - y1)^2)
Point-slope equation of a line: y - y1 = m(x - x1)
Completing the square

Theorems

A point lies on a curve if it satisfies the equation of the curve
Perpendicular distance from a point to a line or curve

Suitable Grade Level

Grades 10-12