1.2 The Tangent Line Problem for Circles and Parabolas

Goals

Before we begin our formal study of calculus, we solve the the tangent line problem in the context of circles and parabolas.

It is possible to find the slope, and hence, an equation of the tangent line to a circle or parabola via algebraic or geometric procedures. Thus, in this section, you will

– review the definition of a circle and the equation of a circle in Cartesian coordinates;

– review the technique of completing the square to find the center and the radius of a circle;

– apply the technique of completing the square to find the vertex of a parabola;

– find equations of lines tangent to a circle at a given point;

– find equations of lines tangent to a parabola at a given point;

– learn a problem-solving technique to find points on a circle given a geometric constraint about their tangent lines;

– apply the above problem-solving technique to find points on a parabola given a geometric constraint about its tangent lines.

Prior Knowledge

 
Distance formula. Let \(P\left (x_1, y_1\right )\) and \(Q\left (x_2, y_2\right )\) be two points on the \(xy\)-plane. Then, the distance} between the points is
\[|\, PQ \,| = \sqrt{\left (x_2-x_1\right )^2+ \left (y_2-y_1\right )^2}.\]

Midpoint. Let \(P\left (x_1, y_1\right )\) and \(Q\left (x_2, y_2\right )\) be two points on the \(xy\)-plane. Then, the midpoint} between these points is
\[M= \left (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right ).\]

Lines. Let \(P\left (x_1, y_1\right )\) and \(Q\left (x_2, y_2\right )\) be two points on the \(xy\)-plane. Then, the slope} of the line \(\mathcal{L}\) passing through these points is
\[m = \frac{y_2-y_1}{x_2-x_1}.\]

Point-slope equation of \(\mathcal{L}\) (this is not unique).
\[y -y_1=m\left (x-x_1\right )\;\;\; \text{or} \;\;\; y -y_2=m\left (x-x_2\right ).\]
The slope-intercept equation of \(\mathcal{L}\) (this is unique):
\[ y = mx + b.\]
Here, \((0,b)\) is the \(y\)-intercept of the line.
 

Assuming that \(m\neq 0\), then, the slope of the line perpendicular to \(\mathcal{L}\) is given by
\[m_{\perp} = -\frac{1}{m}.\]

1.2.1 Circles and Parabolas

Definition of a Circle

Definition 1. A circle is a set of points in a plane equidistant from a fixed point.
The fixed point is called the center of the circle, and the common distance the radius of the circle.
 

Example 1. Let \(\mathcal{C}\) be the circle with center \(C(1,-2)\) and radius \(2\).

(a) Show that the point \(A(1+\sqrt{3}, -1)\) lies on \(\mathcal{C}\).

(b) Find the points on the circle \(\mathcal{C}\) whose \(x\)-coordinate is \(\displaystyle \frac{3}{2}\).

Solution. (a) We need to show that \(d(A,C)=2\), that is,
\[d(A,C)=\sqrt{(1-(1+\sqrt{3}))^2+(-2-(-1))^2} = \sqrt{3+1}=2.\]

(b) Here we are looking for the points \(P\left(\frac{3}{2},y\right)\) such that \(d(P,C)=2\), that is,

\begin{align*}
d(P,C)=\sqrt{\left(1-\frac{3}{2}\right)^2+(-2-y)^2} =2
&\underset{?}{\Longleftrightarrow} \sqrt{\frac{1}{4}+(y+2)^2}=2
\Longleftrightarrow \frac{1}{4}+(y+2)^2=4 \Longleftrightarrow
\\
&\underset{?}{\Longleftrightarrow} (y+2)^2=\frac{15}{4}
\underset{?}{\Longleftrightarrow} y = -2\pm\frac{\sqrt{15}}{2}
\end{align*}

Thus, the points are \(\displaystyle \left(\frac{3}{2},-2 – \frac{\sqrt{15}}{2}\right)\) and \(\displaystyle \left(\frac{3}{2},-2 + \frac{\sqrt{15}}{2}\right)\).

Equation of a Circle in Cartesian Coordinates

Let \(\mathcal{C}\) denote the set of points on the circle with center at \(C(h,k)\) and radius \(r\). Then
\[P(x,y) \in \mathcal{C} \Longleftrightarrow d(P(x,y),C(h,k))= r \Longleftrightarrow\sqrt{(x-h)^2+(y-k)^2}= r\]
or, by squaring both sides,
\begin{equation}\label{ch01-eq03}
(x-h)^2+(y-k)^2=r^2. \tag{1.3}
\end{equation}
Example 2. Given the points \(A(1,2)\) and \(B(-3, 5)\), find an equation for the circle that has as a diameter the line segment \(AB\).

Solution. (a) First, it is necessary to determine the center and the radius of the circle.
The center is the midpoint of the line segment \(AB\). Thus,
\[C(h,k)=\left(\frac{1+(-3)}{2}, \frac{2+5}{2}\right)= \left(-1, \frac{7}{2}\right).\]
The radius is half of the diameter, and the diameter is precisely the length of the segment \(AB\). Thus,
\[r = \frac{1}{2}\sqrt{\left(-3-1\right)^2 + \left(5-2\right)^2}
=\frac{1}{2}\sqrt{16 + 9}=\frac{5}{2}.\]
Conclusion. The equation of the circle is given by
\[ \left(x-(-1)\right)^2+\left(y-\frac{7}{2}\right)^2=\left(\frac{5}{2}\right)^2
\;\; \text{or}\;\;\left(x+1\right)^2+\left(y-\frac{7}{2}\right)^2=\frac{25}{4}.\]

Finding the Center and the Radius of a Circle

Very often, an equation for a circle is given in expanded form:
\(Ax^2+Ay^2+Dx + Ey +F=0\) (same coefficient for \(x^2\) and \(y^2\)). In this case, the technique used to find the center and the radius is by completing the square. Central to this technique are the following factorization formulas:
\[(a+b)^2 = a^2+2ab + b ^2\hspace{.75in}(a-b)^2 = a^2-2ab + b ^2.\]
Here is the sequence of steps to carry out when completing the square:

(1) Make the coefficients of \(x^2\) and \(y^2\) equal to 1.

(2) Group terms in \(x\) and terms in \(y\). Constant terms on the other side of the equation.

(3) Rewrite the terms in \(x\) and \(y\) in a way to match either one of the expressions
\[(a+ b)^2=a^2+ 2ab + b^2\;\;\; \text{or}\;\;\; (a – b)^2=a^2 – 2ab + b^2.\]
Note that you must add the same quantity to both sides of the equation.

(4) Identify each squared binomial.

Example 3. Show that the equation \(2x^2+2y^2+12x-8y-24=0\) is the equation of a circle, and find its center and its radius.

Solution.
\[2x^2+2y^2+12x-8y-24=0
\overset{(1)}{\Longleftrightarrow} x^2+y^2+6x-4y-12=0 \Longleftrightarrow\]
\[ \overset{(2)}{\Longleftrightarrow} x^2+6x+y^2-4y=12
\overset{(3)}{\Longleftrightarrow} \left(x^2+2\cdot x\cdot 3+3^2\right)+\left(y^2-2\cdot y \cdot 2+2^2\right)=12+3^2+2^2 \Longleftrightarrow\]
\[\overset{(4)}{\Longleftrightarrow} (x+3)^2+\left(y-2\right)^2=12+9+4=25 \Longrightarrow C\left(-3,2\right),\;\;
r=\sqrt{25}=5.
\]

Finding the Vertex and the \(x\)- and \(y\)-Intercepts of a Parabola

 
The technique of completing the square can be applied to find the vertex of a parabola. This is how it can be done in general. Let \(y = f(x) = ax^2 + bx + c\), \(a \neq 0\), be a quadratic function, then
\begin{align*}
y &= ax^2 + bx + c = a \left( x^2 + \frac{b}{a}x + \frac{c}{a} \right)
= a\left( {\underbrace {x^2 + 2x\frac{b}{2a} + \left( {\frac{b}{{2a}}} \right)^2}_{u^2 + 2uv + v^2} – \left( \frac{b}{2a} \right)^2 + \frac{c}{a}} \right) = \\
&= a\left( {\underbrace {{{\left( {x + \frac{b}{{2a}}} \right)}^2}}_{{{(u + v)}^2}} – \frac{{{b^2}}}{{4a^2}} + \frac{c}{a}} \right) = a\left( {{{\left( {x + \frac{b}{{2a}}} \right)}^2} + \frac{{4ac – {b^2}}}{{4a^2}}} \right) = a{\left( {x + \frac{b}{{2a}}} \right)^2} + \frac{{4ac – {b^2}}}{4a}.
\end{align*}
There are two cases to consider:

\(a > 0\). In this case,
\[ y= a{\left( {x + \frac{b}{{2a}}} \right)^2} + \frac{{4ac – {b^2}}}{4a} \geq \frac{{4ac – {b^2}}}{4a},\]
and the parabola opens upward. The vertex is the lowest point on the graph of the parabola and has coordinates
\[
\hspace{.5in}\hphantom{(*)}
V\left (-\frac{b}{2a}, f\left (-\frac{b}{2a}\right )\right ) = \left (-\frac{b}{2a}, \frac{{4ac – {b^2}}}{4a}\right ).
\hspace{.5in}(*)
\]

\(a < 0\). In this case,
\[ y= a{\left( {x + \frac{b}{{2a}}} \right)^2} + \frac{{4ac – {b^2}}}{4a} \leq \frac{{4ac – {b^2}}}{4a},\]
and the parabola opens downward. The vertex is the highest point on the graph of the parabola and has coordinates as in \((*)\) above.
 

Example 4. Given the parabola \(y=2x^2-3x+1\), find its (a) vertex; (b) \(x\)- and \(y\)-intercepts; (c) sketch its graph.

Solution. (a) Completing the square, we have
\begin{align*}
y &= 2 x^2-3 x+1
= 2\left ( x^2-\frac{3}{2} x+\frac{1}{2}\right )
= 2\left ( x^2-\frac{3}{2} x+ \left (\frac{3}{4}\right )^2 – \left (\frac{3}{4}\right )^2 + \frac{1}{2}\right )=\\
&=2\left ( \left (x -\frac{3}{4}\right )^2 – \frac{9}{16} + \frac{1}{2}\right )= 2\left (x -\frac{3}{4}\right )^2 – \frac{1}{8}. \;\;\Longrightarrow \;\; V\left (\frac{3}{4}, -\frac{1}{8}\right ).
\end{align*}

(b) There are two ways to proceed to find the \(x\)-intercepts:

\item[(i)] \(\displaystyle f(x)=0\;\;\Longleftrightarrow \;\; 2 x^2-3 x+1 = 0 \;\;\Longleftrightarrow \;\; (x-1) (2 x-1)= 0 \;\;\Longleftrightarrow \;\;x\in \left \{\frac{1}{2}, 1\right \}. \)

Thus, the \(x\)-intercepts are given by \(\displaystyle \left \{\left (\frac{1}{2}, 0\right ), \left (1, 0\right )\right \}\)
\item[(ii)] Use the expression for \(f\) after completing the square in (a).
\begin{align*}
f(x)= 0 &\;\; \Longleftrightarrow\;\;2\left (x -\frac{3}{4}\right )^2 – \frac{1}{8} =0 \;\; \Longleftrightarrow\;\;\left (x -\frac{3}{4}\right )^2 =\frac{1}{16} \;\; \Longleftrightarrow\;\; x -\frac{3}{4} =\pm \frac{1}{4} \;\;\Longleftrightarrow\\
&\;\; \Longleftrightarrow\;\; x =\frac{3}{4} +\pm \frac{1}{4} \;\;\Longleftrightarrow \;\; x\in \left \{\frac{1}{2}, 1\right \}. \;\; \Longrightarrow \;\; \left \{\left (\frac{1}{2}, 0\right ), \left (1, 0\right )\right \}.
\end{align*}

\(y\)-intercept: \((0, f(0)) = (0,1)\).

(c) Note that since \(y=2x^2-3x+1\), then \(a=2 > 0\), and the parabola opens upward.



Figure 1.3: Example 4

1.2.2 Tangent Lines to Circles

Note. We are going to use (without proof) the fact that the tangent line to a circle at a point \(A\) is perpendicular to the radial line passing through \(A\) (see Figure 1.3).

Example 1. Given the circle \(\mathcal{C}: \, x^2+2y^2+12x-8y-24=0\),(a) verify that the point \(A(1,5)\) lies on the circle; (b) find the slope-intercept equation of the line tangent to \(\mathcal{C}\) at \(A\).

Solution. From Example 3 in 1.2.1, we know that \(C\left(-3,2\right)\) and \(r=\sqrt{25}=5\).

(a)
\[d(C,A)=\sqrt{(1-(-3))^2+(5-2)^2} =\sqrt{16+9}=5.\]
(b) Given that the tangent line to the circle at \(A\) is perpendicular to the radial line passing through \(A\) (see Figure 1.4), we compute the slope of the radial line:
\[ m_0=\frac{5-2}{1-(-3)}=\frac{3}{4}.\]

Figure 1.4: Example 1

It follows that the tangent line has slope
\[m=-\frac{1}{m_0}=-\frac{4}{3}.\]
Hence, the point-slope equation of the tangent line is given by
\[y-5=-\frac{4}{3}(x-1),\]
The slope-intercept equation is given by
\[y=-\frac{4}{3}x+\frac{19}{3}.\]

1.2.4 Problem Solving: Tangent Lines to Circles and Parabolas

Tangent Lines to Circles

Example 1. Find the points on \(x^2+y^2-2x+6y=15\) whose tangent lines pass through the point \(\displaystyle A\left(-\frac{22}{3},-3\right)\).

Solution. Figure 1.5 shows that there are two such points.

Let \(P(x,y)\) be one such point on the circle. In order to find its coordinates, we need to set up two equations (because there are two unknowns).

One equation is readily available:

Since \(P(x,y)\) is on the circle, it must satisfy the equation of the circle
\begin{equation}\label{ch01-eq04}
x^2+y^2-2x+6y=15 \;\;\; \text{or} \;\;\; (x-1)^2+(y+3)^2=5^2. \tag{1.4}
\end{equation}

Figure 1.5: Example 1

To obtain a second equation, we compute the slope of the line segment \(AP\) in two different ways:

(i) Use the definition of the slope of a line. Using the coordinates \(\displaystyle A\left(-\frac{22}{3},-3\right)\) and \(P(x,y)\), the slope is \(\displaystyle m=\frac{y+3}{\displaystyle x+\frac{22}{3}}\).

(ii) Use the fact that the radial line \(PC\) is perpendicular \(AP\).

First we need to determine the center and the radius of the circle. Using Equation (\ref{ch01-eq04}), we have that the center is \(C(1,-3)\) and the radius is \(r=5\).

Then, the slope of the radial line \(CP\) is given by \(\displaystyle m_{CP}=\frac{y+3}{x-1}\), and, because \(AP \perp PC\), we have a second expression for \(m\),
\[m = -\frac{1}{\displaystyle \frac{y+3}{x-1}} = -\frac{x-1}{y+3},\;\; x\neq 1.\]

Since we have found two different ways to represent \(m\), these expressions have to be equal. Namely,
\begin{equation}\label{ch01-eq05}
\frac{y+3}{\displaystyle x+\frac{22}{3}} = -\frac{x-1}{y+3}.\tag{1.5}
\end{equation}
Thus, to find the coordinates of \(P(x,y)\), it is necessary to solve equations (\ref{ch01-eq04}) and (\ref{ch01-eq05}) simultaneously.

Equation (\ref{ch01-eq03}) can be rewritten as follows: assuming that \(\displaystyle x\neq -\frac{22}{3}\) and \(y\neq -3\), then
\[\frac{y+3}{\displaystyle x+\frac{22}{3}} = -\frac{x-1}{y+3}
\Longleftrightarrow
(y+3)^2 = -\left (x+\frac{22}{3}\right )(x-1)
\Longleftrightarrow
x^2 + y^2 + \frac{19}{3}x + 6y = -\frac{5}{3}.
\]
Thus, we need to solve simultaneously the equations
\begin{equation}\label{ch01-eq06}
(x-1)^2+(y+3)^2=5^2\;\;\text{ and }\;\;(y+3)^2 = -\left (x+\frac{22}{3}\right )(x-1). \tag{1.6}
\end{equation}
Note that the conditions \(\displaystyle x\ne -\frac{22}{3}\) and \(y\ne -3\) are no longer necessary in (\ref{ch01-eq06}) because these are the coordinates of the point \(A\), and \(A\) is outside the circle. Thus, the equation of the circle rules out these values.

Solving simultaneously (\ref{ch01-eq06}). Note that we can eliminate \((y+3)^2\) to get a single equation for \(x\). Rewriting (\ref{ch01-eq06}), we have
\[
(y+3)^2=5^2-(x-1)^2 \;\;\; \text{and}\;\;\; (y+3)^2 = -\left(x+\frac{22}{3}\right)(x-1).\]
Thus
\[5^2 – (x-1)^2 = -\left(x+\frac{22}{3}\right)(x-1)
\Longleftrightarrow
25 = (x-1)^2-\left(x+\frac{22}{3}\right)(x-1)
\Longleftrightarrow
\]
\[
\Longleftrightarrow
25=(x-1)\left((x-1)-\left(x+\frac{22}{3}\right)\right)
\Longleftrightarrow 25=(x-1)\left(-\frac{25}{3}\right)
\underset{?}{\Longleftrightarrow}
x=-2.\]
To find the corresponding \(y\)-coordinates, substitute \(x=-2\) into either one of the two equations in (\ref{ch01-eq06}):
\[(-2-1)^2+(y+3)^2=5^2 \Longleftrightarrow (y+3)^2 =16 \Longleftrightarrow y = -3\pm 4
\Longleftrightarrow y=1 \text{ or } y=-7.\]
Conclusion. The points on the circle whose tangent lines pass through \(\displaystyle A\left(-\frac{22}{3},-3\right)\) are \((-2,1)\) and \((-2, -7)\).

Tangent Lines to Parabolas

Example 2. Given the parabola \(\mathcal{P}\) with equation \(y=2 x^2-4 x+3\), and the secant line \(\mathcal{L}_S\) passing through the points \(A\) and \(B\) with \(x\)-coordinates, respectively, \(x=0\) and \(x=4\),

(a) find the slope-intercept equation of the line \(\mathcal{L}_T\) tangent to \(\mathcal{P}\) that is parallel to \(\mathcal{L}_S\);

(b) find the point \(P(x,y)\) where \(\mathcal{L}_T\) is tangent to \(\mathcal{P}\).

Intuitive Idea of the Solution. Figure 1.6 shows (a) the parabola and the secant line; (b) several lines parallel to the secant line. We can think of these lines as sliding up and down the secant line. Intuitively, at some point one such line will be tangent to the parabola as in (c).

Figure 1.6: (a)

(b)

(c)

Solution. Figure 1.6 (c) shows the graph of the parabola, the secant line \(\mathcal{L}_S\), and the tangent line \(\mathcal{L}_T\) parallel to \(\mathcal{L}_S\). \(\mathcal{L}_S\) passes through the points with coordinates
\[A(0, f(0))=(0,3)\;\; \text{and } \;\; B(4, f(4))= (4, 19) \;\;\text{(verify this).}\]
Thus, \(\mathcal{L}_S\) has slope
\[m_S=\frac{19-3}{4-0}=4,\]
which is also the slope of \(\mathcal{L}_T\) (since \(\mathcal{L}_S \, \parallel\, \mathcal{L}_T\)).

Thus, to find the slope-intercept equation for \(\mathcal{L}_T\), \(y=m x +b\), we only need to find \(b\), since \(m=4\), that is, \(y = 4x +b\).

To find \(b\), note that \(\mathcal{L}_T\), being tangent to the parabola, intersects the parabola at the single point \(P(x,y)\). Thus, we are looking for the value of \(b\) such that
\[y = 4x +b\;\; \text{and } \;\; y=2 x^2-4 x+3\]
intersect at a single point. Solving simultaneously,
\[\hspace{.3in} \hphantom{(*)} 4x +b = 2 x^2-4 x+3 \Longleftrightarrow 2x^2-8x+3-b=0 \Longleftrightarrow x=\frac{8\pm \sqrt{64-8(3-b)}}{4}. \hspace{.3in} (*)\]
Thus, the graphs intersect at a single point if and only if (why)
\[64-8(3-b)=0 \Longleftrightarrow b=-5.\]
Hence, the slope-intercept equation of \(\mathcal{L}_T\) is \( y = 4x-5\).

And, using the equation \((*)\), it follows that the \(x\)-coordinate of the point where \(\mathcal{L}_T\) is tangent to \(\mathcal{P}\) is \(\displaystyle x=\frac{8}{4}=2\).
Thus, its \(y\)-coordinate is \(y=4(2)-5 = 3\) (why).

Conclusion. The line tangent to the parabola and parallel to \(\mathcal{L}_S\) has slope-intercept equation \(y=4x-5\) and is tangent to the parabola at the point \(P(2,3)\).

Example 3. Given the parabola \(\mathcal{P}:\) \(\displaystyle y=\frac{1}{4} x^2- x+2\), and the point \(A(3,-1)\), find the points on the parabola whose tangent lines pass through \(A\).

Figure 1.7: Example 3

Solution. Figure 1.7 shows the parabola and the two points whose tangent lines pass through \(A\). Let \(P(x_0,y_0)\) be one such point. In order to find its coordinates, we need to set up two equations.

One equation is the equation of the parabola itself—as the point \(P(x_0,y_0)\) lies on the parabola:
\[\hspace{1in} \hphantom{(*)} y=\frac{1}{4} x^2- x+2. \hspace{1in}(*)\]

A second equation will be obtained from the condition that the line through \(A\) and \(P\) must be tangent to the parabola.

Before we set up such an equation, we first determine the slopes of the lines passing through \(A\) and tangent to the parabola. Let \(m\) denote the slope of one such line. Then
\[\hspace{1in} \hphantom{(**)}y-(-1)= m (x-3) \;\; \Longleftrightarrow \;\; y = mx – 3m -1\hspace{1in}(**)\]
is the equation of the line. This line must intersect the parabola at a single point. Thus, when solving, simultaneously, \((*)\) and \((**)\), there must be a unique solution (for \(x\)). We have
\begin{align*}
\hspace{.04in} \hphantom{(***)}\frac{1}{4} x^2- x+2 = mx – 3m -1& \;\; \Longleftrightarrow \;\; \frac{1}{4} x^2- (m+1)x++ 3m + 3 = 0 \:\:\Longleftrightarrow \\
&
\:\:\Longleftrightarrow\;\;x = \frac{\displaystyle m+1\pm \sqrt{(m+1)^2-4\cdot\frac{1}{4}\cdot (3m + 3)}}{\displaystyle 2\cdot\frac{1}{4}}\hspace{.19in} (***)
\end{align*}
It follows that there is a unique solution for \(x\) if and only if the expression inside the radical is \(0\), that is,
\[\Longleftrightarrow \;\; (m+1)^2-3(m + 1) = 0 \;\; \Longleftrightarrow \;\; (m+1)(m-2)=0 \;\; \Longleftrightarrow \;\; m \in \left \{-1,2\right \}.\]
Hence, the tangent lines to the parabola passing through \(A(3,-1)\) have equations given by
\[y = -x+2\hspace{.5in} y = 2x -7.\]
Furthermore,
\[\text{if } \; m=-1 \;\; \underset{(***)}{\Longrightarrow}\;\; x=0, \;\;\;\; \text{ and if } \; m=2 \;\; \underset{(***)}{\Longrightarrow}\;\; x=6.\]
Thus, the \(x\)-coordinates of the points of tangency are \(0\) and \(6\). And the \(y\)-coordinates are obtained as follows:
\[\text{if } \; x=0,\; m=-1 \;\; \underset{(**)}{\Longrightarrow}\;\; y=2, \;\;\;\; \text{ and if } \; x=6, \;m=2 \;\; \underset{(**)}{\Longrightarrow}\;\; y=5.\]
Therefor, the points of tangency are \(\left \{(0,2), (6,5)\right \}\).