Math Problem Statement

1. The line y = 2x - 8 meets the coordinate axes at A and B. The line AB is a diameter of the circle. Find the equation of the circle. 2. The circle center (8, 10) meets the x-axis at (4, 0) and (a, 0). a) Find the radius of the circle. b) Find the value of a.

Solution

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Question 1:

"The line y=2x8y = 2x - 8 meets the coordinate axes at points AA and BB. The line ABAB is a diameter of the circle. Find the equation of the circle."

  • Step 1: Find intercepts (A and B)

    • When the line meets the x-axis, y=0y = 0: 0=2x8    x=40 = 2x - 8 \implies x = 4 So, point A=(4,0)A = (4, 0).
    • When the line meets the y-axis, x=0x = 0: y=2(0)8=8y = 2(0) - 8 = -8 So, point B=(0,8)B = (0, -8).
  • Step 2: Find the center and radius of the circle

    • The center of the circle is the midpoint of the diameter ABAB: Midpoint of A(4,0) and B(0,8)=(4+02,0+(8)2)=(2,4)\text{Midpoint of } A(4, 0) \text{ and } B(0, -8) = \left( \frac{4 + 0}{2}, \frac{0 + (-8)}{2} \right) = (2, -4)
    • The radius is half the distance of ABAB: Distance AB=(40)2+(0(8))2=16+64=80=45\text{Distance } AB = \sqrt{(4 - 0)^2 + (0 - (-8))^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5} So, the radius r=452=25r = \frac{4\sqrt{5}}{2} = 2\sqrt{5}.
  • Step 3: Equation of the circle The general 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.

    • Center (2,4)(2, -4)
    • Radius r=25r = 2\sqrt{5}, so r2=20r^2 = 20.

    The equation of the circle is: (x2)2+(y+4)2=20(x - 2)^2 + (y + 4)^2 = 20

Question 2:

"The circle center (8,10)(8, 10) meets the x-axis at (4,0)(4, 0) and (a,0)(a, 0)."

  • Part (a): Find the radius of the circle.

    • The radius is the distance from the center (8,10)(8, 10) to any point on the circle, such as (4,0)(4, 0). r=(84)2+(100)2=16+100=116r = \sqrt{(8 - 4)^2 + (10 - 0)^2} = \sqrt{16 + 100} = \sqrt{116} So, the radius is 116\sqrt{116}, which simplifies to 2292\sqrt{29}.
  • Part (b): Find the value of aa.

    • The radius must also be the distance from the center (8,10)(8, 10) to (a,0)(a, 0). We know the radius is 116\sqrt{116}, so: (8a)2+(100)2=116\sqrt{(8 - a)^2 + (10 - 0)^2} = \sqrt{116} Squaring both sides: (8a)2+100=116    (8a)2=16    8a=±4(8 - a)^2 + 100 = 116 \implies (8 - a)^2 = 16 \implies 8 - a = \pm 4 Solving this: a=84=4ora=8+4=12a = 8 - 4 = 4 \quad \text{or} \quad a = 8 + 4 = 12 Therefore, a=12a = 12 (since we already have a=4a = 4 as one solution).

Would you like further details on any of these solutions? Here are some questions to expand on this topic:

  1. How can we derive the equation of a line from two points?
  2. What other ways can we calculate the radius of a circle from its diameter?
  3. Why does a circle's equation take the form (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2?
  4. Can you generalize how to find the intercepts of a line for any linear equation y=mx+cy = mx + c?
  5. How do you verify if a given point lies on a circle?

Tip: Always check the units of radius and other distances in geometry problems to ensure they are consistent when plugging into formulas.

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Math Problem Analysis

Mathematical Concepts

Geometry
Analytic Geometry
Circle Equations
Linear Equations

Formulas

Midpoint formula: Midpoint = ((x1 + x2)/2, (y1 + y2)/2)
Distance formula: Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Circle equation: (x - h)^2 + (y - k)^2 = r^2

Theorems

Diameter Theorem
Properties of Circles
Distance between two points in a plane

Suitable Grade Level

Grades 10-12