Math Problem Statement

Write the standard equation of the circle with the given properties: 1. C(2,3) and radius is 4 units 2. Center at the origin and (3, -4) is a point on the circle 3. Endpoints of the diameter are (-1,2) and (5,6).

Solution

Let's go through the problems one by one and find the standard equation of the circles for each case.

1. Given: C(2,3)C(2,3) and radius r=4r = 4

The standard form of a circle's equation is:

(xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2

Where (h,k)(h, k) is the center and rr is the radius.

Here, the center CC is at (2,3)(2,3) and the radius r=4r = 4.

So, the equation becomes:

(x2)2+(y3)2=42(x - 2)^2 + (y - 3)^2 = 4^2

(x2)2+(y3)2=16(x - 2)^2 + (y - 3)^2 = 16

2. Given: Center at the origin (0,0)(0,0) and (3,4)(3, -4) is a point on the circle.

First, calculate the radius using the distance formula between the center and the point on the circle:

r=(x2x1)2+(y2y1)2r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

r=(30)2+(40)2=9+16=25=5r = \sqrt{(3 - 0)^2 + (-4 - 0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5

So, the equation of the circle is:

x2+y2=52x^2 + y^2 = 5^2

x2+y2=25x^2 + y^2 = 25

3. Given: Endpoints of the diameter are (1,2)(-1, 2) and (5,6)(5, 6).

First, find the center, which is the midpoint of the diameter:

Midpoint=(1+52,2+62)=(42,82)=(2,4)\text{Midpoint} = \left( \frac{-1 + 5}{2}, \frac{2 + 6}{2} \right) = \left( \frac{4}{2}, \frac{8}{2} \right) = (2, 4)

Next, find the radius, which is half the distance between the endpoints:

Distance=(x2x1)2+(y2y1)2\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Distance=(5(1))2+(62)2=62+42=36+16=52\text{Distance} = \sqrt{(5 - (-1))^2 + (6 - 2)^2} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52}

Radius r=522=4×132=2132=13r = \frac{\sqrt{52}}{2} = \frac{\sqrt{4 \times 13}}{2} = \frac{2\sqrt{13}}{2} = \sqrt{13}.

So, the equation is:

(x2)2+(y4)2=(13)2(x - 2)^2 + (y - 4)^2 = (\sqrt{13})^2

(x2)2+(y4)2=13(x - 2)^2 + (y - 4)^2 = 13


Summary of Equations:

  1. (x2)2+(y3)2=16(x - 2)^2 + (y - 3)^2 = 16
  2. x2+y2=25x^2 + y^2 = 25
  3. (x2)2+(y4)2=13(x - 2)^2 + (y - 4)^2 = 13

Would you like further details or have any questions?

Relative Questions:

  1. How do you derive the distance formula used to calculate the radius?
  2. What are the steps to find the midpoint between two points?
  3. How do you generalize the equation of a circle for a given center and radius?
  4. Can the equation of a circle be derived using the Pythagorean theorem? If so, how?
  5. How would the equation change if the circle was translated to a different quadrant?

Tip:

Always double-check your midpoint and distance calculations to ensure accuracy when determining the circle's equation.

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Math Problem Analysis

Mathematical Concepts

Geometry
Coordinate Geometry
Circles

Formulas

Standard form of a circle's equation: (x - h)^2 + (y - k)^2 = r^2
Distance formula: r = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Midpoint formula: Midpoint = ((x1 + x2)/2, (y1 + y2)/2)

Theorems

Theorem of a Circle's Equation
Distance Theorem

Suitable Grade Level

Grades 9-11