MHB Equation of a line tangent to a circle

Click For Summary
To find the value of m for the circle x^2 + y^2 - 4x + 2y + m = 0 to be tangent to the line y = x + 1, substitute y into the circle's equation to form a quadratic in x. The discriminant of this quadratic, given by b^2 - 4ac, must equal zero for the line to be tangent, indicating one solution. Completing the square reveals the circle's center at C(2, -1) and establishes that the radius is perpendicular to the tangent line. By determining the intersection point and calculating the distance from the center to this point, the value of m can be derived.
Armela
Messages
1
Reaction score
0
The circle x^2 +y^2 -4x+2y+m=0 is tangent with the line y=x+1.Find m.

p.s : I know that o should solve it from the equations of two lines but i really get confused when i substitute the y :/ .
Thanx :)
 
Mathematics news on Phys.org
Hi Armela,

Welcome to MHB! :)

Geometry is not my strong suit at all, but I did find http://planetmath.org/EquationOfTangentOfCircle.html for you which might be useful.

Jameson
 
Armela said:
The circle x^2 +y^2 -4x+2y+m=0 is tangent with the line y=x+1.Find m.

p.s : I know that o should solve it from the equations of two lines but i really get confused when i substitute the y :/ .
Thanx :)

When you substitute $y=x+1$ in the equation of the circle, you get a quadratic equation for $x$. Solve that quadratic equation using the "$\sqrt{b^2-4ac}$" formula (the solution will involve the constant $m$). If $b^2-4ac$ is positive then there are two solutions to the equation, meaning that the line cuts the circle in two points. If it is negative then there are no solutions, meaning that the line misses the circle. But if it is zero then there is just one (repeated) solution, meaning that the line is tangent to the circle.
 
Armela said:
The circle x^2 +y^2 -4x+2y+m=0 is tangent with the line y=x+1.Find m.

p.s : I know that o should solve it from the equations of two lines but i really get confused when i substitute the y :/ .
Thanx :)


Maybe I understand your remark about the two lines completely wrong, but here comes a way to use actually two lines:

1. Determine the center of the circle by completing the squares. You should come out with:

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

So the center is at C(2, -1)

2. If the given line is a tangent to the circle then the radius of the circle is perpendicular to the given line at the tangent point T.
The given line has the slope m = 1 therefore the line frome the center C to the tangent point T has the slope m = -1.
Determine the equation of the line CT. You should come out with

$y = -x+1$

3. Determine the intercept between the given line and CT to get the coordinates of T. You should come out with $T(0, 1)$.

4. Calculate the distance $r=|\overline{CT}|$.

Since $r^2=5-m$ you are able to determine the value of m.
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
7
Views
2K