Concentration Core v1

Circle Maker: Visual Proof

Don't memorize \((x-h)^2 + (y-k)^2 = r^2\). Just watch it happen.

1. The Visual Proof

Move the sliders to see how the center \((h, k)\) and radius \(r\) transform the circle.

Area (πr²)
50.3
Circumference (2πr)
25.1
x y r=4

Practice Challenge General Form of a Circle

Find the radius of the circle in the xy-plane with the following equation:

\[x^2 + y^2 + 4x - 6y - 3 = 0\]
(A) 2
(B) 3
(C) 4
(D) 5
Click to see the "Desmos Speed-Run"

Quick Strategy

1. Type the equation exactly into Desmos to see the circle.

2. Draw a horizontal line passing through the center (origin) of the circle. Since the center is at \((-2, 3)\), type y = 3.

Desmos Step 1: Center Line

3. To find the diameter, find a line starting from one edge of the circle to the other edge. Restrict the domain: y = 3 {-6 ≤ x ≤ 2}.
(Tip: To type ≤ in Desmos, type < followed immediately by =)

Desmos Step 2: Diameter Segment

4. From this we get the diameter (\(-6\) to \(2\) is \(8\)).

5. The radius is just a half of the diameter, so the radius is \(4\).

Formula Check:
\( r = \sqrt{\frac{D^2}{4} + \frac{E^2}{4} - F} = \sqrt{\frac{4^2}{4} + \frac{(-6)^2}{4} - (-3)} = \sqrt{\frac{16}{4} + \frac{36}{4} + 3} = \sqrt{4 + 9 + 3} = \sqrt{16} = 4 \)
Wait, where does this formula come from?

It comes from matching coefficients between the Standard Form and the General Form.

  1. Standard Form: \((x - h)^2 + (y - k)^2 = r^2\)
  2. Expand it: \(x^2 - 2hx + h^2 + y^2 - 2ky + k^2 - r^2 = 0\)
  3. Rearrange powers: \(x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0\)
  4. General Form is: \(x^2 + y^2 + Dx + Ey + F = 0\)
  5. Match the pieces:
    • \(D = -2h \Rightarrow h = -\frac{D}{2}\)
    • \(E = -2k \Rightarrow k = -\frac{E}{2}\)
    • \(F = h^2 + k^2 - r^2\)

If you isolate \(r^2\) from the \(F\) equation:

\(r^2 = h^2 + k^2 - F\)

Substitute the \(h\) and \(k\) fractions:

\(r^2 = \left(-\frac{D}{2}\right)^2 + \left(-\frac{E}{2}\right)^2 - F\)

\(r^2 = \frac{D^2}{4} + \frac{E^2}{4} - F\)

Square root both sides:

\(r = \sqrt{\frac{D^2}{4} + \frac{E^2}{4} - F}\)

In our problem, \(x^2 + y^2 + 4x - 6y - 3 = 0\), so \(D=4\), \(E=-6\), and \(F=-3\).

Hard Challenge Circle-Line Intersection

A circle in the xy-plane has equation \((x-3)^2 + (y+1)^2 = 25\). If a line has equation \(y = 3\), at what two x-coordinates does the line intersect the circle?

(A) \(-1\) and \(7\)
(B) \(0\) and \(6\)
(C) \(1\) and \(5\)
(D) \(-2\) and \(8\)
Click to see the Solutions

Method 1: The Desmos Speed-Run

1. Type (x-3)^2 + (y+1)^2 = 25

2. Type y = 3

3. Click the gray intersection dots!

Desmos Circle Line Intersection

Desmos shows (0, 3) and (6, 3) instantly. The answer is (B).

Method 2: Algebraic Substitution

Substitute \(y = 3\) into the circle's equation:

\((x - 3)^2 + (3 + 1)^2 = 25\)
\((x - 3)^2 + 4^2 = 25\)
\((x - 3)^2 + 16 = 25\)
\((x - 3)^2 = 9\)

Take the square root of both sides (remember to include the \(\pm\)):

\(x - 3 = \pm 3\)
\(x = 3 \pm 3\)

This gives us two solutions for \(x\):

  • \(x = 3 - 3 = 0\)
  • \(x = 3 + 3 = 6\)

The two x-coordinates are 0 and 6. The answer is (B).