A geometry question

Question

A circle is given by the equation x^2-6x+y^2+2y=15. Write the equation of the tangent line, where (6, 3) is the point of tangency.

I saw this in a math textbook, and it looked kind of challenging. (For me).

Answer 1

At an advanced level, you might not even deal with the fact that this is a circle. However, as this is asked with the title geometry, I am solving it in a geometric manner.

First, we know that it is a circle because x^2 and y^2 have a coefficient of 1. Of course, the constant on the right side must be greater than 0 when we get done.

First, we complete the square to get into center-radius form.

x^2 - 6x + 9 - 9 + y^2 + 2y +1 - 1 = 15

(x-3)^2 + (y+1)^2 = 15 + 9 + 1

(x-3)^2 + (y+1)^2 = 5^2

Our center is at (3, -1) and the radius is 5.

Now, we are asked about the tangent at (6, 3)

We check to see if this point is on the circle by subtracting the center from it and calculating the distance. This also tells us the slope of the tangent line, which is perpendicular to the line from the center to the point.

(6, 3) - (3, -1) = (3, 4) √(3^2+4^2) = √25 = 5. The point is on the circumference of the circle.

Since the slope of the line from the center to this point is 4/3, the slope of the tangent line, which is the negative reciprocal, is -3/4

Then, we can use y = mx + b, plugging in the point (6, 3), which is on the line

6 = -3/4 (3) + b

b + 6 + 3/4(3) = 8 1/4

y = -3/4 x + 8 1/4 is the tangent line.

If you wish, you may remove fractions

4y = -3x + 33,

and we reorganize as

3x + 4y - 33 = 0

Note that the coefficients of x and y could have been determined from the direction of the point from the center, (3, 4), but that would be getting ahead of ourselves.