Math Problem Statement

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The image contains five different circle-related math questions involving equations of circles, finding centers, radii, and tangents.

Solution

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Question 1

The circle CC has the equation: x2+y22x+6y=26x^2 + y^2 - 2x + 6y = 26 We need to:

  1. Find the coordinates of the center of C.
  2. Find the radius of C.

Step 1: Rearranging the equation into standard form

We will complete the square for both xx and yy.

  • For xx, we have x22xx^2 - 2x. Completing the square: x22x=(x1)21x^2 - 2x = (x - 1)^2 - 1

  • For yy, we have y2+6yy^2 + 6y. Completing the square: y2+6y=(y+3)29y^2 + 6y = (y + 3)^2 - 9

Now, rewrite the equation with these squared terms: (x1)21+(y+3)29=26(x - 1)^2 - 1 + (y + 3)^2 - 9 = 26 Simplify: (x1)2+(y+3)2=26+1+9(x - 1)^2 + (y + 3)^2 = 26 + 1 + 9 (x1)2+(y+3)2=36(x - 1)^2 + (y + 3)^2 = 36

Step 2: Finding the center and radius

  • The standard form of a circle 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.
  • Comparing this with (x1)2+(y+3)2=36(x - 1)^2 + (y + 3)^2 = 36, we can see that:
    • The center of the circle is (1,3)(1, -3).
    • The radius rr is 36=6\sqrt{36} = 6.

Answer:

  • The coordinates of the center are (1,3)(1, -3).
  • The radius is 66.

Question 2

The circle CC has center (2,5)(2, 5) and passes through the point (4,9)(4, 9). Find the equation for CC.

Step 1: Use the formula of a circle

The standard equation of a circle 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.

Step 2: Finding the radius

We are given the center (2,5)(2, 5) and a point on the circle (4,9)(4, 9). The radius is the distance between these two points: r=(42)2+(95)2=22+42=4+16=20=25r = \sqrt{(4 - 2)^2 + (9 - 5)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}

Step 3: Writing the equation of the circle

The equation of the circle is: (x2)2+(y5)2=(25)2(x - 2)^2 + (y - 5)^2 = (2\sqrt{5})^2 (x2)2+(y5)2=20(x - 2)^2 + (y - 5)^2 = 20

Answer: The equation of the circle is (x2)2+(y5)2=20(x - 2)^2 + (y - 5)^2 = 20.


Question 3

The circle CC has center (2,3)(-2, 3) and passes through point (1,8)(1, 8).

  1. (a) Find the equation for C.

Step 1: Use the formula of a circle

The center is (2,3)(-2, 3) and the radius can be found by calculating the distance from (2,3)(-2, 3) to (1,8)(1, 8): r=(1(2))2+(83)2=32+52=9+25=34r = \sqrt{(1 - (-2))^2 + (8 - 3)^2} = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}

Step 2: Writing the equation of the circle

The equation of the circle is: (x+2)2+(y3)2=(34)2(x + 2)^2 + (y - 3)^2 = (\sqrt{34})^2 (x+2)2+(y3)2=34(x + 2)^2 + (y - 3)^2 = 34

Answer: The equation of the circle is (x+2)2+(y3)2=34(x + 2)^2 + (y - 3)^2 = 34.

  1. (b) Show that the point (3,6)(3, 6) lies on C.

Substitute x=3x = 3 and y=6y = 6 into the equation: (3+2)2+(63)2=34(3 + 2)^2 + (6 - 3)^2 = 34 52+32=25+9=345^2 + 3^2 = 25 + 9 = 34 Since both sides are equal, the point (3,6)(3, 6) lies on the circle.


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Math Problem Analysis

Mathematical Concepts

Geometry of Circles
Coordinate Geometry
Equations of Circles
Tangent to a Circle
Distance Formula

Formulas

Equation of a circle: (x - h)^2 + (y - k)^2 = r^2
Distance formula: d = √((x2 - x1)^2 + (y2 - y1)^2)
Tangent Line to a Circle Formula

Theorems

Equation of a Circle in Standard Form
Distance Between Two Points Theorem
Tangent to a Circle Theorem

Suitable Grade Level

Grades 10-12