Showing Tangency of a Line and Circle: Simplifying the Equation Method

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To determine if a line is tangent to a circle, one effective method is to rearrange both equations and substitute to find intersection points. If the resulting quadratic equation has a discriminant of zero, it indicates that the line intersects the circle at exactly one point, confirming tangency. The discussion also highlights that if the length of the perpendicular from the circle's center to the line equals the circle's radius, the line is tangent. An example is provided where the equations x^2 + y^2 - 10y = 0 and 4x - 3y + 40 = 0 are analyzed, demonstrating that they meet at a single point with matching slopes. This confirms that the line is indeed tangent to the circle.
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If i have the equation of a circle and the equation of a line how do i show if the line is a tangent of the circle?

At first i tried rearranging them both in the form f(x) = 0, setting them to equal each other. Rearranging and simplifying, then eliminating x or y by using the linear line equation. I was then going to try to find the points at which they meet and hence differentiate both equations to find dy/dx at this point - if they were equal i would have shown it was a tangent.
However, this usually doesn't work. I'm not sure if there's something wrong with my algebraic manipulation but it's rarely solvable with the quadratic equation.

Is there an easier way? If so, what is it?
 
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Length of perendicular from centre of circle should be qual to radius of circle for the line to be tanget.
other way ut the equation of circle in form of one variable by using equation of line and show discriminant is zero of quadratic equation so formed. rovig line intersects at only one point.
 
Thanks for the help, would this be an accepable method?

Prove that x^2 + y^2 - 10y = 0 has a tangent 4x - 3y + 40 = 0.

Rearrange to form 4x = 3y-40, x = 3/4 y -10

Substitute into the equation of the circle:

(3/4 y - 10)^2 + y^2 - 10y = 0
Simplifies to y^2 -16y + 64 = 0

Solves to give y = only 8, hence at this point x = -4.

If x^2 + y^2 - 10y = 0
dy/dx = -2x/(2y-10)
Substitute in the x and y gives dy/dx = 8/6 = 4/3

Of the equation of the tangent line:
4x = 3y - 40
dy/dx = 4/3.

They meet at only one point and at this point they have the same gradient, hence they are tangents.
 
Gaz,
When u say,
"...Substitute into the equation of the circle..." (fourth line)

What are u doing here ??...
u are finding the intersection points of the line with the circle.

Now every line usually will intersect in two points with the circle but our given line intersects at one and only one point.

So by definition of tangent, the given line is tangent to the circle.

So the following working,
"If x^2 + y^2 - 10y = 0
dy/dx = -2x/(2y-10)
Substitute in the x and y gives dy/dx = 8/6 = 4/3

Of the equation of the tangent line:
4x = 3y - 40
dy/dx = 4/3."

is not necessary.

Essentially, u have solved the problem in complete. :)

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