A calculus of the absurd
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10.2 Circles
All the points which are a common distance from a single point collectively form a circle. How far is every point on a circle from the centre? The radius! If we have two points \(q\) and \(p\) in space, we can draw a right-angled triangle between them, and obtain that the distance
between these two points is given by the formula
\(\seteqnumber{0}{10.}{1}\)
\begin{equation}
\norm {-c + q} = \sqrt {(p_x - q_x)^2 + (p_y - q_y)^2}
\end{equation}
which is true because of Pythagoras’ theorem ^{66}^{66} This was explained further in the section above..
A circle can be thought of as all the points which are the same distance (i.e. the radius) from a common point (i.e. the centre).
Thus in general we can write the equation of a circle which has a centre at \((c_x, c_y)\) and radius \(r\) as
\(\seteqnumber{0}{10.}{2}\)
\begin{equation}
\label {general circle equation with square root} \sqrt { (x - c_x)^2 + (y - c_y)^2 } = r
\end{equation}
and square both sides to obtain that
\(\seteqnumber{0}{10.}{3}\)
\begin{equation}
\label {general circle equation squared} (x - c_x)^2 + (y - c_y)^2 = r^2
\end{equation}
Solution: The first step here is to find the minimum point of \(y=x^2-2x+b\), which can be done by completing the square.
\(\seteqnumber{0}{10.}{4}\)
\begin{equation}
x^2 - 2x + b = (x-1)^2 + b - 1
\end{equation}
Looking at the above equation, we know that the minimum point is when \(y\) is at its smallest possible value. This is when \((x-1)^2\) is at its smallest possible value (which is zero, as \(x^2 \geqq 0\), and thus zero is the smallest value it can be), which is when \(x=1\) and
thus \((x-1)^2=0\). Any other value of \(x\) would give a bigger value. When \(x=1\), we have \(y=b-1\) and thus we have found the minimum point \((1, b-1)\).
If you recall that the equation of a circle essentially specifies a distance from the centre, then our circle is just saying that any point on the circle is exactly \(\sqrt {20}\) units from the centre. We’re not interested in points on the circle, though, we’re
interested in any points within the circle. These are any points where the distance between \((1, b-1)\) and the centre of the circle \((2, -3)\) is less than \(\sqrt {20}\). Thus we can write down an inequality.
\(\seteqnumber{0}{10.}{5}\)
\begin{equation}
\sqrt { (1 - 2)^2 + ((b-1) - (-3))^2 } < \sqrt {20}
\end{equation}
The left-hand side is just Pythagoras’ theorem (sketch the two points in space, and then draw a right-angled triangle between them). With this inequality, we can then start to manipulate, starting by squaring both sides. Usually it’s dangerous to square both sides of an inequality, as
negative numbers become positive, and thus break the whole thing. However, in this equation all the values must be positive (as adding two squares is just adding two positive numbers, which will always be positive, and \(\sqrt {20}>0\)) so it’s fine. Thus,
\(\seteqnumber{0}{10.}{6}\)
\begin{align*}
& (1 - 2)^2 + ((b-1) - (-3))^2 < 20 \\ & 1 + (b+2)^2 < 20 \\ & b^2 + 4b + 4 + 1 < 20 \\ & b^2 + 4b - 15 < 0 \\
\end{align*}
If we sketch the quadratic, we get
After using the quadratic formula to find the roots, we can tell that the minimum point of the curve is within the circle for all values of \(b\) which are in the range \(-6.359<b<2.359\).