Q. 52 - Precalculus Enhanced with Graphing Utilities

Question

The tangent line to a circle may be defined as the line that intersects the circle in a single point, called the point of tangency. See the figure. 

If the equation of the circle is $x^2+y^2=r^2$ and the equation of the tangent line is y = mx + b, show that:

(a) $r^2(1+m^2)=b^2$ 

[Hint: The quadratic equation $x^2+{(mx+b)}^2=r^2$ has exactly one solution.]

(b) The point of tangency is $(\frac{-r^{2}m}{b},\frac{r^2}{b})$.

(c) The tangent line is perpendicular to the line containing the center of the circle and the point of tangency.

Step-by-Step Solution

Answer

The following have been shown in detail in the solution:

(a) $r^2(1+m^2)=b^2$

(b) The point of tangency is $(\frac{-r^{2}m}{b},\frac{r^2}{b})$.

(c) The tangent line is perpendicular to the line containing the center of the circle and the point of tangency.

Part (a) Step 1. Given information

Given that the equation of the circle is $x^2+y^2=r^2$ and the equation of the tangent line is y = mx + b.

We have to show that $r^2(1+m^2)=b^2$.

Part (a) Step 2. Solve the system of equations

Replace y=mx+b in the equation of the circle.

$$\begin{gathered} x^2+{(mx+b)}^2=r^2 \\ {x}^2+m^2{x}^2+b^2+2mbx=r^2 \\ (1+m^2)x^2+2mbx+b^2=r^2 \\ (1+m^2)x^2+2mbx+b^2-r^2=0 \\ (1+m^2)x^2+\left(2mb\right)x+(b^2-r^2)=0 \end{gathered}$$

A quadratic equation has only one solution if its discriminant is zero.

Part (a) Step 3. Put discriminant to zero.

Discriminant of a quadratic equation $a^2+bx+c=0$ is $b^2-4ac$.

Therefore, the discriminant of $(1+m^2)x^2+\left(2mb\right)x+(b^2-r^2)=0$ will be:

${\left(2mb\right)}^2-4(a+m^2)(b^2-r^2)$.

$$\begin{gathered} {\left(2mb\right)}^2-4(a+m^2)(b^2-r^2)=0 \\ \begin{array}{c}\left(4m^2{b}^2\right)-4\left(1+m^2\right)b^2+4\left(1+m^2\right)r^2=0\\ 4m^2{b}^2-4b^2-4m^2{b}^2+4\left(1+m^2\right)r^2=0\\ 4\left(1+m^2\right)r^2=4b^2\\ \left(1+m^2\right)r^2=b^2\end{array} \end{gathered}$$

Hence proved.

Part (b) Step 1. Solve the system of equations

Replace y=mx+b in the equation of the circle.

$$\begin{gathered} x^2+{(mx+b)}^2=r^2 \\ {x}^2+m^2{x}^2+b^2+2mbx=r^2 \\ (1+m^2)x^2+2mbx+b^2=r^2 \\ (1+m^2)x^2+2mbx+b^2-r^2=0 \\ (1+m^2)x^2+\left(2mb\right)x+(b^2-r^2)=0 \end{gathered}$$

Part (b) Step2. Use the quadratic formula x = - b ± b 2 - 4 a c 2 a

Using the quadratic formula for $(1+m^2)x^2+\left(2mb\right)x+(b^2-r^2)=0$.

$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ =\frac{-b\pm \sqrt{D}}{2a} \\ =\frac{-b}{2a} (\mathrm{Since} \mathrm{D} \mathrm{is} \mathrm{zero}) \\ =\frac{-\left(2\mathrm{mb}\right)}{2\left(1+{\mathrm{m}}^2\right)} \\ =\frac{-\mathrm{mb}}{1+{\mathrm{m}}^2} \\ \mathrm{Replace} 1+{\mathrm{m}}^2 \mathrm{with} \frac{{\mathrm{b}}^2}{{\mathrm{r}}^2} \mathrm{from} \mathrm{part} \left(\mathrm{a}\right) \\ \mathrm{x}=\frac{{\displaystyle -\mathrm{mb}}}{{\displaystyle \frac{{\displaystyle {\mathrm{b}}^2}}{{\displaystyle {\mathrm{r}}^2}}}}=-\frac{{\mathrm{r}}^2\mathrm{m}}{\mathrm{b}} \\ \mathrm{Substitute} \mathrm{this} \mathrm{value} \mathrm{in} \mathrm{y}=\mathrm{mx}+\mathrm{b} \\ \mathrm{y}=\mathrm{m}(-\frac{{\displaystyle {\mathrm{r}}^2\mathrm{m}}}{{\displaystyle \mathrm{b}}})+\mathrm{b} \\ =\frac{-{\mathrm{r}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}{\mathrm{b}} \\ \mathrm{Substitute} {\mathrm{b}}^2={\mathrm{r}}^2(1+{\mathrm{m}}^2) \mathrm{from} \mathrm{part} \left(\mathrm{a}\right) \\ \mathrm{y}=\frac{{\displaystyle -{\mathrm{r}}^2{\mathrm{m}}^2+{\mathrm{r}}^2(1+{\mathrm{m}}^2)}}{{\displaystyle \mathrm{b}}} \\ =\frac{{\displaystyle -{\mathrm{r}}^2{\mathrm{m}}^2+{\mathrm{r}}^2+{\mathrm{r}}^2{\mathrm{m}}^2}}{{\displaystyle \mathrm{b}}} \\ =\frac{{\mathrm{r}}^2}{\mathrm{b}} \\ \mathrm{Hence} \mathrm{the} \mathrm{point} \mathrm{of} \mathrm{tangency} \mathrm{is} (-\frac{{\displaystyle {\mathrm{r}}^2\mathrm{m}}}{{\displaystyle \mathrm{b}}},\frac{{\displaystyle {\mathrm{r}}^2}}{{\displaystyle \mathrm{b}}}). \end{gathered}$$

Part (c) Step 1. Find the slope of the line joining origin and point of tangency

Coordinates of origin are $(0,0)$.

Coordinates of point of tangency are $(-\frac{{\displaystyle {\mathrm{r}}^2\mathrm{m}}}{{\displaystyle \mathrm{b}}},\frac{{\displaystyle {\mathrm{r}}^2}}{{\displaystyle \mathrm{b}}})$.

Therefore, the slope of the line joining origin and point of tangency will be:

$\frac{{\displaystyle \frac{r^2}{b}}-0}{{\displaystyle \frac{-r^{2}m}{b}}-0}=-\frac{1}{m}$

Part (c) Step 2. Find the product of slopes of the tangent line and the line joining origin and point of tangency

Slope of the tangent line is m.

Slope of the line joining origin and point of tangency is $-\frac{1}{m}$.

The product is .$m\times (-\frac{1}{m})=-1$

Hence the two lines are perpendicular.