Tangent: Slope Form (Part 2)


This lesson will talk about a few things related to the previous lesson – the point of contact of the tangent of given slope, and the equation of the tangent in the slope form, when the center of the circle is not at origin.

Let’s go back to the equation which we derived in the previous lesson: y = mx ± a\(\sqrt{1+m^2}\) . Now I’m interested in the point where this tangent will touch the circle x2 + y2 = a2.

The point of contact will be the root of the quadratic equation which we had obtained by solving the circle and the line, i.e. (1 + m2)x2 + 2cmx + c2 – a2­ = 0.

Recall that if quadratic equation ax2 + bx + c = 0 has equal roots, or the discriminant equals 0, then the both roots will be equal to -b/2a.

Using this, the x-coordinate of the point of contact will be -2cm/(1 + m2). By multiplying and dividing by a2 and putting a2(1 + m2) = c2 (the condition for tangency, obtained in the previous lesson), the final expression obtained for the x-coordinate is -a2m/c

To find the y-coordinate, I’ll simply plug in the value of x in the line y = mx + c. The obtained value will be a2/c (I suggest you do the calculations yourself)

We have the coordinates of the point of contact (-a2m/c, a2/c), where c = ± a\(\sqrt{1+m^2}\)

 

Moving on, let’s now derive the equation of the tangent for the circle, whose center is not at origin.

Consider the general equation: x2 + y2 + 2gx + 2fy + c = 0, which can be written in the form (x + g)2 + (y + f)2 = r2, where r2 = g2 + f2 – c

To find the tangent’s equation in this case, I’ll use a small trick – assume the line to be (y + f) = m(x + g) + k, instead of y = mx + k.

Why? Because, on replacing x + g by X and y + f by Y, in the equations of the line and the circle, the calculations involved and the results obtained will exactly be the same.

That is, on solving the equations X2 + Y2 = r2 and Y = mX + k, and putting D = 0, we’ll get k = ± r\(\sqrt{1+m^2}\)

So the equation of the tangent becomes: (y + f) = m(x + g) ± r\(\sqrt{1+m^2}\)

What about the point of contact? Exactly the same: (–r2m/k, r2/k) where k = ± r\(\sqrt{1+m^2}\) , except for one small thing: here X = – r2m/k and Y = r2/k. So x = – g – r2m/k and y = – f + r2/k

The trick involved here was something known as “Shifting of Origin”, which I haven’t yet covered in my lessons. Please feel free to explore the topic on your own. I’ll be posting related lessons later on.

To summarize:

  1. The point of contact of the tangent of slope ‘m’ with the circle x2 + y2 = a2 is (-a2m/c, a2/c), where c = ± a\(\sqrt{1+m^2}\)
  2. Equation of the tangent of slope ‘m’, to the circle x2 + y2 + 2gx + 2fy + c = 0 is given by (y + f) = m(x + g) ± r \(\sqrt{1+m^2}\), where r is the radius of the circle.

Also the point of contact of the tangent will be (– g – r2m/k, – f + r2/k) where k = ± r\(\sqrt{1+m^2}\)

Note that in both the cases, there will be two tangents for a given value of m, hence two points of contact in each case.

That’s it for the current lesson. Time for some examples!

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