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Geometry

Find Circle Through Three Points

Determine the equation of a circle passing through three given points using completing the square method and algebraic techniques.

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Problem

Find the equation of the circle passing through the three points A=(0,0)A=(0,0), B=(6,0)B=(6,0), and C=(0,8)C=(0,8).

Step 1: Use the general circle form

Write the circle as

x2+y2+dx+ey+f=0.x^2+y^2+dx+ey+f=0.

Since it passes through A=(0,0)A=(0,0), substitution gives f=0f=0.

Step 2: Substitute the other two points

Using B=(6,0)B=(6,0):

36+6d=0,36+6d=0,

so

d=6.d=-6.

Using C=(0,8)C=(0,8):

64+8e=0,64+8e=0,

so

e=8.e=-8.

That leaves

x26x+y28y=0.x^2-6x+y^2-8y=0.

Step 3: Complete the square

Complete the square in both variables by adding 99 and 1616:

x26x+9+y28y+16=25.x^2-6x+9+y^2-8y+16=25.

This becomes

(x3)2+(y4)2=25.(x-3)^2+(y-4)^2=25.

So the center is (3,4)(3,4) and the radius is 55.

Step 4: Check with the geometric shortcut

Since A\angle A is a right angle, BCBC is the diameter of the circle. With AB=6AB=6 and AC=8AC=8, the Pythagorean theorem gives

BC=36+64=10,BC=\sqrt{36+64}=10,

so the radius is 55 and the midpoint of BCBC is (3,4)(3,4), matching the algebraic result.

Answer

The circle is

(x3)2+(y4)2=25.(x-3)^2+(y-4)^2=25.

Concepts

Equations of Circles

The standard equation of a circle with center (h,k)(h, k) and radius rr is (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2. A general form x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0 can be converted to standard form by completing the square.

Equations with Variables on Both Sides

Linear equations where the unknown appears on both sides of the equal sign, such as 3x+2=x+103x + 2 = x + 10. To solve, collect all variable terms on one side and all constant terms on the other, then simplify to find the value of the variable.

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