# Circle and tangents

• Jul 16th 2009, 01:35 AM
dhiab
Circle and tangents
• Jul 16th 2009, 05:02 AM
earboth
Quote:

Originally Posted by dhiab

1. All lines with the equation $y = m(x-4)$ pass through the point P(4,0).

2. Calculate the coordinates of the points of intersection between all lines and the circle. If the line is a tangent then there exists only one point of intersection, that means the discriminante must be zero:

Plug in the linear term for y into the 1st equation:

$x^2-2x+(m(x-4))^2-4(m(x-4)) = 0$

$x^2·(m^2 + 1) - x·(8·m^2 + 4·m + 2) + 16·m^2 + 16·m = 0$

Solve for x:

$x = - \dfrac{\sqrt{- 4·m^2 - 12·m + 1} - 4·m^2 - 2·m - 1}{m^2 + 1} \vee$ $x = \dfrac{\sqrt{- 4·m^2 - 12·m + 1} + 4·m^2 + 2·m + 1}{m^2 + 1}$

As stated before the discriminant must be zero:

$-4m^2-12m+1=0~\implies~m=-\dfrac32-\dfrac12 \sqrt{10}~\vee~m=-\dfrac32+\dfrac12 \sqrt{10}$

3. Plug in this value into the equation of the line.

I've attached a sketch of the circle and the 2 tangents.
• Jul 16th 2009, 05:04 AM
Opalg
Substitute $y+m(x-4)$ in the equation for the circle. You will get a quadratic equation for x. Write down the condition for this quadratic to have equal roots. That will give you an equation for m.
• Jul 16th 2009, 05:32 AM
galactus
I have another way if that's OK.

By completing the square, we find the circle has equation

$(x-1)^{2}+(y-2)^{2}=5$

The tangent line will be a distance from the center of the circle equal to the radius length of $\sqrt{5}$. The center of the circle is at (1,2)

Using the distance between a line and a point formula:

$\frac{|-m(1)+1(2)+4m|}{\sqrt{(-m)^{2}+1}}=\sqrt{5}$

Solving for m gives $m=\frac{\sqrt{10}-3}{2}, \;\ \frac{-(\sqrt{10}+3)}{2}$
• Jul 16th 2009, 05:34 AM
Soroban
Hello, dhiab!

I think I understand what you're asking . . .

Quote:

$C$ is the circle: . $x^2 + y^2 - 2x - 4y \:=\:0$

A tangent to $C$ is: . $y \:=\:m(x-4)$

$\text{Find }m.$

The circle is: . $(x-1)^2 + (y-2)^2 \:=\:5$

. . It has center $P(1,2)$ and radius $r = \sqrt{5}$

The line . $y - 0 \:=\:m(x-4)$ .has slope $m$ and point $Q(0,4)$

$Q$ is one of the $y$-intercepts of the circle.

We are asked for the slope of the tangent at $Q.$

Code:

            |  /             | /  * * *         Q  |/*          *       (0,4)o              *           *| *              *           / |  *         /* |    * P        *         / * |      o        *           * |    (1,2)      *             |           *|                *       - - - * - - - - - - - * - - -             | *          *             |    * * *             |

The radius $PQ$ has slope: . $\frac{4-2}{0-1} \:=\:-2$

The tangent at $Q$ is perpendicular to that radius.

Therefore: . $m \:=\:\frac{1}{2}$

• Jul 16th 2009, 06:45 AM
dhiab
Quote:

Originally Posted by Soroban
Hello, dhiab!

I think I understand what you're asking . . .

The circle is: . $(x-1)^2 + (y-2)^2 \:=\:5$

. . It has center $P(1,2)$ and radius $r = \sqrt{5}$

The line . $y - 0 \:=\:m(x-4)$ .has slope $m$ and point $Q(0,4)$

$Q$ is one of the $y$-intercepts of the circle.

We are asked for the slope of the tangent at $Q.$

Code:

            |  /             | /  * * *         Q  |/*          *       (0,4)o              *           *| *              *           / |  *         /* |    * P        *         / * |      o        *           * |    (1,2)      *             |           *|                *       - - - * - - - - - - - * - - -             | *          *             |    * * *             |
The radius $PQ$ has slope: . $\frac{4-2}{0-1} \:=\:-2$

The tangent at $Q$ is perpendicular to that radius.

Therefore: . $m \:=\:\frac{1}{2}$

Hello : THANK YOU
All lines with the equation http://www.mathhelpforum.com/math-he...7a429670-1.gif pass through the point P(4,0)AND NOT TANGENT IN THE POINT.