Thread: 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 ) .

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 $\displaystyle y = m(x-4)$ 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:

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

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

Solve for x:

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

As stated before the discriminant must be zero:

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

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

I've attached a sketch of the circle and the 2 tangents.

Substitute $\displaystyle y+m(x-4)$ 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.

I have another way if that's OK.

By completing the square, we find the circle has equation

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

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

Using the distance between a line and a point formula:

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

Solving for m gives $\displaystyle m=\frac{\sqrt{10}-3}{2}, \;\ \frac{-(\sqrt{10}+3)}{2}$

Hello, dhiab!

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

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

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

$\displaystyle \text{Find }m.$

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

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


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

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


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

Code:

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

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

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

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

Originally Posted by Soroban

Hello, dhiab!

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

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

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


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

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


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

Code:

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

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

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

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

Hello : THANK YOU
All lines with the equation pass through the point P(4,0)AND NOT TANGENT IN THE POINT.