Question

1.  An archeologist found the remains of an ancient wheel, which she then placed on a grid. If an arc of the wheel passes through A(-7,0), B(-3,4) and C(7,0), located the center of the wheel, and the standard equation of the circle defining its boundary. 

Answer 1

The standard equation of the circle defining its boundary is

$$(x-a)^2 + (y-b)^2 = R^2$$

where the point with the coordinates (a, b) is the center of the circle. Put each point in the equation. Then we get the system of equations:

$$\left\{ \begin{array}{rcl} {(-7-a)^2 + (0 - b)^2 = R^2,} \\ {(-3-a)^2 + (4 - b)^2 = R^2,} \\ {(7-a)^2 + (0 - b)^2 = R^2.} \end{array} \right.$$

Open brackets (use the formula $(a+b)^2 = a^2 + b^2 + 2ab$) :

$$\left\{ \begin{array}{rcl} {49 +a^2+ 14a + b^2 = R^2,} \\ {9+a^2 +6a+ 16+ b^2 - 8b = R^2,} \\ {49 +a^2- 14a + b^2 = R^2.} \end{array} \right.$$

Subtract from the first equation the third:

$49 +a^2+ 14a + b^2 - (49 +a^2- 14a + b^2) = R^2 -R^2$

$28a = 0 \implies a = 0$

Equate the first equation to the second:

$49 +a^2+ 14a + b^2 = 9+a^2 +6a+ 16+ b^2 - 8b$

Because $a = 0$:

$49 +b^2 = 9+16+ b^2 - 8b$

$25 - 8b = 49 -8b = 24 \implies b = -3$

Then the center of the circle is (0, -3).

Find the radius of the circle: put point and the coordinates of center in the equation:

$(-7-0)^2 + (0 - (-3))^2 = R^2$

$49 +9 = R^2 \implies R =\sqrt{58}$

The standard equation of the circle:

$$x^2 + (y+3)^2 = 58$$