Question

the equation of the tangents to the circle x^2+y^2-2rx-2by+b^2=0 from origin

Answer 1

We expect the tangents are of the form y = mx + c. However, since they pass through the origin, c = 0. Hence, the tangents are expected to be of the form y = mx. (There are surprises to come though! This is quite a long and detailed answer; read it carefully!)

By definition, tangents are lines which only meet the circle once. So, our task here is to determine the value(s) of m for which y = mx intersects the circle exactly once.

Well, we begin by attempting to find where y = mx intersects the circle.

Substitute y = mx into the given equation of the circle, to eliminate y:

$x^{2} + y^{2} - 2rx - 2by + b^{2} = 0$

$\iff x^{2} + (mx)^{2} - 2rx - 2b(mx) + b^{2} = 0$

$\iff x^{2} + m^{2}x^{2} - 2rx - 2bmx + b^{2} = 0$

$\iff (1 + m^{2})x^{2} - 2(r - bm)x + b^{2} = 0$

Now, this is what we call our "solution quadratic" for finding the x-coordinates of the points of intersection. We know however, that our y = mx only intersects the circle once. This means that our "solution quadratic" above has only one solution. This means that its discriminant is equal to zero.

So, y = mx is a tangent to the circle ⇔ discriminant of above quadratic = 0

$\iff [2(r-bm)]^{2} - 4(1 + m^{2})b^{2} = 0$

$\iff 4(r^{2} - 2rbm  + b^{2}m^{2}) - 4b^{2} - 4m^{2}b^{2} = 0$

$\iff 4r^{2} - 8rbm + 4b^{2}m^{2} - 4b^{2} - 4m^{2}b^{2} = 0$

$\iff 8rbm = 4(r^{2} - b^{2})$

$\iff m = \frac{r^{2} - b^{2}}{2rb}.$

[For some reason an A with a hat on top appears in the above calculation. Please ignore this!]

Of course that last step up there is only valid if neither r nor b is zero. Since you haven't mentioned whether or not this is the case in the question, we must consider all the possibilities. Also, one other thing to note is that we only got one value for m and hence one tangent (for the case r ≠ 0, b ≠ 0). This is quite strange isn't it? Now we know that the origin doesn't lie strictly inside the circle (if it did there would be no tangents through the origin). So what exactly is going on? Where's the other value?

Well, to sort all the issues raised in the above paragraph, we will write the equation of the circle in completed square form:

$x^{2} + y^{2} - 2rx - 2by + b^{2} = 0$

$\iff x^{2} - 2rx + y^{2} - 2by = -b^{2}$

$\iff (x - r)^{2} - r^{2} + (y - b)^{2} - b^{2} = -b^{2}$

$\iff (x - r)^{2} + (y - b)^{2} = r^{2}.$

From this form, we deduce immediately that the centre of the circle is (r, b) and the radius is r. We can use these to discuss what would happen if either r or b were to be zero.

If r = 0, then radius of circle is equal to zero and this is not possible, so we conclude that in fact, r ≠ 0.

If b = 0 and r ≠ 0, then the centre of the circle is (r, 0). In this case, we note that the circle passes through the origin (draw a picture maybe?) ,and part of the x-axis is a radius. So, in this case, since the y-axis is perpendicular to the x-axis, and the circle passes through the origin, the only tangent through the origin is x = 0 (the y -axis).

If b ≠ 0 and r ≠ 0, then one of the tangents is what we found above by using the discriminant. This tangent is :

$y = \left(\frac{r^{2} - b^{2}}{2rb}\right)x.$

However, the last question we asked was where is the other tangent in this case? We know there are two tangents because here origin is neither inside the circle nor on the circle (substitute x = 0, y = 0 to equation of circle and check!). So where is the other tangent? To answer this, we note that the centre of the circle is (r, b) and the radius is r. This means that the circle meets the y-axis only at (0, b) (again draw a diagram!!!). We can actually show this algebraically:

Substitute x = 0 to equation of circle:

$(0 - r)^{2} + (y - b)^{2} = r^{2}$

$\iff (y - b)^{2} = 0$

$\iff y - b = 0$

$\iff y = b.$

What does this mean? Well, it means that since the circle only meets the y-axis at one point, the y-axis must also be a tangent to the circle.

Now you might wonder why this didn't come out when we set discriminant to zero earlier. The answer to that is that the equation of the y-axis is actually not of the form y = mx. In fact, the y-axis has infinite gradient which is why it didn't come up from the discriminant method.

To summarise:

r = 0 is not possible because then the circle would have radius zero.

If r ≠ 0 and b = 0, then the only tangent to the circle through the origin is the line x = 0 (also known as the y-axis).

If r ≠ 0 and b ≠ 0 then there are two tangents through the origin, whose equations are:

$\underline{\underline{y = \left(\frac{r^{2} - b^{2}}{2rb}\right)x \ and\   x = 0}}.$

[For some reason, an A with a hat on top appears in the answer above. If you see this, then please ignore it as it shouldn't be there!]

[This answer has been edited a few times to add in notes about A hats appearing in places where they shouldn't exist. I think this is a bug with the Latex.]