Chapter 1.4, Problem 52AYU

(a)

To determine

To show that $r^2\left(1+m^2\right)=b^2$ .

Explanation of Solution

Given information:

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

Concept Used:

• The tangent line to a circle is the line that intersects the circle at a single point, which is known as point of tangency.

Calculation:

Consider the equation of the circle,

  $x^2+y^2=r^2\mathrm{......}\left(i\right)$

And the equation of the tangent line

  $y=mx+b\mathrm{......}\left(ii\right)$

Substitute the value of y from ( ii ) into ( i ) and simplify further as shown below,

  $\begin{array}{c}{x}^2+y^2=r^2\\ x^2+{\left(mx+b\right)}^2=r^2\\ x^2+m^2{x}^2+2mxb+b^2=r^2\\ x^2\left(1+m^2\right)+\left(2mb\right)x+\left(b^2-r^2\right)=0\mathrm{......}\left(iii\right)\end{array}$

Now solve it using the quadratic formula to get

  $\begin{array}{c}x=\frac{-\left(2mb\right)\pm \sqrt{{\left(2mb\right)}^2-4\left(1+m^2\right)\left(b^2-r^2\right)}}{2\left(1+m^2\right)}\\ =\frac{-2mb\pm \sqrt{4m^2{b}^2-4b^2+4r^2-4m^2{b}^2+4m^2{r}^2}}{2\left(1+m^2\right)}\\ =\frac{-2mb\pm \sqrt{-4b^2+4r^2+4m^2{r}^2}}{2\left(1+m^2\right)}\\ =\frac{-2mb\pm 2\sqrt{m^2{r}^2-b^2+r^2}}{2\left(1+m^2\right)}\\ =\frac{-mb\pm \sqrt{m^2{r}^2-b^2+r^2}}{\left(1+m^2\right)}\\ =\frac{-mb}{1+m^2}\left(\text{for real solutions, discriminant part is zero}\right)\end{array}$

Substitute this value of x in ( iii ) to get

  $\begin{array}{c}{x}^2+m^2{x}^2+2mxb+b^2=r^2\\ \left(1+m^2\right){\left(\frac{-mb}{1+m^2}\right)}^2+2mb\left(\frac{-mb}{1+m^2}\right)+b^2=r^2\\ \frac{m^2{b}^2}{1+m^2}-\frac{2m^2{b}^2}{1+m^2}+b^2=r^2\\ \frac{m^2{b}^2-2m^2{b}^2+b^2\left(1+m^2\right)}{1+m^2}=r^2\\ m^2{b}^2-2m^2{b}^2+b^2\left(1+m^2\right)=r^2\left(1+m^2\right)\\ m^2{b}^2-2m^2{b}^2+b^2+b^2{m}^2=r^2\left(1+m^2\right)\\ b^2=r^2\left(1+m^2\right)\end{array}$

It implies that $r^2\left(1+m^2\right)=b^2$ .

(b)

To determine

To show that $r^2\left(1+m^2\right)=b^2$ .

Explanation of Solution

Given information:

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

Concept Used:

• The tangent line to a circle is the line that intersects the circle at a single point, which is known as point of tangency.

Calculation:

The x -coordinate of the point of tangency is given by $x=\frac{-mb}{1+m^2}$ .

From the relation found in part (a), substitute $1+m^2=\frac{b^2}{r^2}$ in above x -coordinate and simplify further as shown below

  $\begin{array}{l}x=\frac{-mb}{\left(\frac{b^2}{r^2}\right)}\\ =-mb\cdot \frac{r^2}{b^2}\\ =\frac{-r^{2}m}{b}\end{array}$

Substitute this value of x in the equation of tangent and simplify further to get

  $\begin{array}{l}y=m\left(\frac{-r^{2}m}{b}\right)+b\\ =\frac{-r^2{m}^2}{b}+b\\ =\frac{-r^2{m}^2+b^2}{b}\\ =\frac{-r^2{m}^2+r^2\left(1+m^2\right)}{b}\\ =\frac{-r^2{m}^2+r^2+r^2{m}^2}{b}\\ =\frac{r^2}{b}\end{array}$

Thus, the point of tangency is $\left(\frac{-r^{2}m}{b},\frac{r^2}{b}\right)$ .

(c)

To determine

To show that $r^2\left(1+m^2\right)=b^2$ .

Explanation of Solution

Given information:

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

Concept Used:

• The tangent line to a circle is the line that intersects the circle at a single point, which is known as point of tangency.

Calculation:

Observe that the center of the circle is at origin (0, 0). So, the slope of the equation of the line connecting the center (0, 0) and the point of tangency $\left(\frac{-r^{2}m}{b},\frac{r^2}{b}\right)$ can be found as

  $\begin{array}{c}\text{Slope, }m=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{\frac{r^2}{b}-0}{\frac{-r^{2}m}{b}-0}\\ =\frac{\frac{r^2}{b}}{\frac{-r^{2}m}{b}}\\ =\frac{r^2}{b}\cdot \frac{b}{-r^{2}m}\\ =-\frac{1}{m}\end{array}$

Also observe that the slope of the tangent line is m .

Since the slope of the line connecting to the center and point of tangency is negative reciprocal of the slope of the tangent line, so both the lines are perpendicular.

Thus, the tangent line is perpendicular to the line containing the center of the circle and the point of tangency.