The Tangent Line

**symmetry** · January 15th 2007, 01:46 PM

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.

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 GIVEN FOR PART A:

The quadratic equation x^2 + (mx + b)^2 = r^2 has exactly ONE SOLUTION.

(b) The point of tangency is the point (-r^2m/b, r^2/b).

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

(d) Draw a tangent line to a circle in the xy-plane.
Use r to represent the radius.

**earboth** · January 15th 2007, 09:58 PM

Originally Posted by symmetry

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.

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 GIVEN FOR PART A:

The quadratic equation x^2 + (mx + b)^2 = r^2 has exactly ONE SOLUTION.

(b) The point of tangency is the point (-r^2m/b, r^2/b).

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

(d) Draw a tangent line to a circle in the xy-plane.
Use r to represent the radius.

Hello,

solve the given quadratic equation for x:

$x^2 + (mx + b)^2 = r^2 \Longleftrightarrow x^2+m^2 x^2 + 2mbx + b^2-r^2=0$ $\Longleftrightarrow (1+m^2)x^2+2mbx+b^2-r^2=0$ . Use the formula to solve a quadratic equation:

$x=\frac{-2mb \pm \sqrt{4m^2 b^2 - 4(1+m^2)(b^2-r^2)}}{2(1+m^2)}$

$x=\frac{-2mb \pm 2 \sqrt{m^2 b^2 -b^2 + r^2 - m^2 b^2 + m^2 r^2}}{2(1+m^2)}$ . There exists only one solution if the radicand equals zero:

$-b^2+r^2+m^2 r^2=0 \Longleftrightarrow r^2(1+m^2)=b^2$

From this equation follows:

$1+m^2=\frac{b^2}{r^2}$

to b)

If the radicand equals zero, than the x-value of the tangent point is

$x=\frac{-mb}{1+m^2}$ . Now use the term to substitute (1+m²) and you'll get:

$x=\frac{-mb}{1+m^2}=\frac{-mb}{\frac{b^2}{r^2}}=\frac{-m r^2}{b}$ . Plug in this term into the equation of the tangent:

$y=\frac{-m r^2}{b} \cdot m + b=\frac{-m^2 r^2 + b^2}{b}$

From the equation of part (a) follows:

$-b^2+r^2+m^2 r^2=0 \Longleftrightarrow r^2=-m^2 r^2 + b^2$ Use this result to simplify the y-value to: $y=\frac{r^2}{b}$

to (c):

The slope of the line from the centre to the tangent point is:

$\frac{y}{x}=\frac{\frac{r^2}{b}}{\frac{-m r^2}{b}}=-\frac{1}{m}$ Two lines are perpendicula if their slopes hold the equation:

$m_2=-\frac{1}{m_1}$ . Thus the tangent and the line containing centre and tangent point are perpendicular.

to (d).

There are 2 different situations:

(i) You already know the tangent point. Then draw the line centre-tangentpoint and afterwards a perpendicular line through the tangent point.

(ii) The tangent passes through a point outside the circle. Then you have to construct a right triangle using the circle of Thales.

EB

**symmetry** · January 16th 2007, 01:44 PM

Absolutely amazing reply with GREAT detail.

Thanks!

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