Equation of a line tangent to a circle

[Armela]

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 :)

 

[Jameson]

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

 

[Opalg]

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.

 

[earboth]

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.

 

[CellCoree]

I would approach this problem using mathematical principles and equations. First, I would rewrite the equation of the circle in standard form, which is (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center of the circle and r is the radius. By completing the square, we can rewrite the given equation as (x-2)^2 + (y+1)^2 = 5+m.

Next, I would use the fact that the line y=x+1 is tangent to the circle to find the point of tangency. This point must lie on both the line and the circle, so we can set the equations equal to each other and solve for x and y. Substituting y=x+1 into the equation of the circle, we get (x-2)^2 + (x+2)^2 = 5+m. Simplifying this equation, we get 2x^2 + 4x + 1 = 5+m.

To find the value of m, we can use the fact that the line is tangent to the circle, which means that the distance from the center of the circle to the point of tangency is equal to the radius. The center of the circle in this case is (2,-1), so we can use the distance formula to set up an equation: √[(x-2)^2 + (x+2)^2] = √(5+m). Solving for x, we get x=1 or x=-3. Since the line is tangent to the circle, we know that the point of tangency must be (1,2) or (-3,-2). Substituting these values into the equation 2x^2 + 4x + 1 = 5+m, we get 5+m = 5 or 5+m = 17. Therefore, m=0 or m=12.

In conclusion, the value of m can be either 0 or 12, depending on the specific point of tangency on the circle. This approach uses the equations of the line and circle to find the point of tangency, and then uses the distance formula to set up an equation to solve for m. This method can be applied to any equation of a line tangent to a circle.