# Thread: Circle and tangents

1. ## Circle and tangents

( C ) the circle for equation :
the straight line for equation : .
( m is the real parametre )
Find m in case : tangent for ( C ) .

2. Originally Posted by dhiab
( C ) the circle for equation :
the straight line for equation : .
( m is the real parametre )
Find m in case : tangent for ( C ) .
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.

3. 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.

4. 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 $m=\frac{\sqrt{10}-3}{2}, \;\ \frac{-(\sqrt{10}+3)}{2}$

5. Hello, dhiab!

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

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

6. 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 pass through the point P(4,0)AND NOT TANGENT IN THE POINT.