Given that the line y=mx+c is a tangent to the circle
(x-a)^2+(y-b)^2 = r^2
show that:
(1-m^2)r^2=(c-b+ma)^2
If the line is tangent, the solution to the system of equations has a single solution.
(x-a)^2 + (y-b)^2 = r^2
(x-a)^2 + (mx+c-b)^2 = r^2
x^2 - 2ax + a^2 + m^2x^2 + 2m(c-b)x + (c-b)^2 = r^2
(m^2+1)x^2 + (2mc-2mb-2a)x + a^2+(c-b)^2-r^2 = 0
To have a single solution, the discriminant must be zero. So, you need
(2mc-2mb-2a)^2 - 4(m^2+1)(a^2+(c-b)^2-r^2) = 0
double-check my math before you try to simplify that monster!
Or, try this way. The tangent line must be perpendicular to the radius which meets it. Also, the distance from the center to the line must be r. Since the center of the circle is at (a,b), if it touches the line at (h,k), we have
m = (a-h)/(mh+c-b)
That must have a single solution, so its discriminant is zero. So,
(c-b)^2 - 4h(b-a) = 0
The distance must equal the radius, so
(h-a)^2 + (mh+c-b)^2 = r^2
Try solving those two equations.
Or google your question. I found several explanations, starting with this one:
https://answers.yahoo.com/question/index?qid=20130528073141AATqi4D
To show that the line y = mx + c is a tangent to the circle (x-a)^2 + (y-b)^2 = r^2, we need to prove that the line intersects the circle at exactly one point.
The first step is to substitute y = mx + c into the equation of the circle, allowing us to find the x-coordinate of the point(s) of intersection. Substituting y = mx + c into the equation, we get:
(x-a)^2 + (mx + c - b)^2 = r^2
Expanding and simplifying this equation, we have:
x^2 - 2ax + a^2 + m^2x^2 + c^2 + b^2 - 2bmx - 2c(m+b)x = r^2
Combining like terms and rearranging, we obtain the following quadratic equation:
(1 + m^2) x^2 + (c - 2am - 2cb)x + (a^2 + b^2 - r^2 + c^2) = 0
For the line to be a tangent to the circle, this quadratic equation must have exactly one real solution for x. In other words, the discriminant of this quadratic equation must be zero.
Setting the discriminant equal to zero, we have:
(c - 2am - 2cb)^2 - 4(1 + m^2)(a^2 + b^2 - r^2 + c^2) = 0
Expanding and simplifying:
c^2 - 4acm - 4bcm - 4c^2b + 4a^2m^2 + 4b^2m^2 - 4m^2r^2 + 4c^2 + 8am^2 + 8bcm + 4a^2 + 4b^2 - 4r^2 = 0
Combining like terms:
(1 - m^2)r^2 + (c - b + ma)^2 = 0
Finally, re-arranging this equation, we obtain the desired result:
(1 - m^2)r^2 = (c - b + ma)^2
Hence, we have shown that (1 - m^2)r^2 = (c - b + ma)^2 when the line y = mx + c is a tangent to the circle (x-a)^2 + (y-b)^2 = r^2.