( 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 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:
Solve for x:
As stated before the discriminant must be zero:
3. Plug in this value into the equation of the line.
I've attached a sketch of the circle and the 2 tangents.
I have another way if that's OK.
By completing the square, we find the circle has equation
The tangent line will be a distance from the center of the circle equal to the radius length of . The center of the circle is at (1,2)
Using the distance between a line and a point formula:
Solving for m gives
Hello, dhiab!
I think I understand what you're asking . . .
The circle is: .is the circle: .
A tangent to is: .
. . It has center and radius
The line . .has slope and point
is one of the -intercepts of the circle.
We are asked for the slope of the tangent at
Code:| / | / * * * Q |/* * (0,4)o * *| * * / | * /* | * P * / * | o * * | (1,2) * | *| * - - - * - - - - - - - * - - - | * * | * * * |
The radius has slope: .
The tangent at is perpendicular to that radius.
Therefore: .