0

Find the equation of the tangent to the circle $$(x-3)^2 + (y-2)^2 = 1$$ at the point $(4,2)$. I cannot show the working of the tangent equation as gradient/slope = 0. So how can I find the gradient of tangent? I think it is also obvious that the tangent is $x = 2$ but what about working?

Matthew Conroy
  • 11,234
Neelkanth
  • 1
  • What have you done so far? You say that is obvious that the tangent is $x=2$, but is there any reason why this is obvious? – Joe Tait Feb 26 '14 at 19:12

3 Answers3

1

The general form for a line is $ax+by = c$. This form more generally describes lines than the form $y=mx+b$, simply the functional implication of the latter form requires us to use a point at infinity in the parameterization of vertical lines.

So, a vertical line at $x=1$ can be described by $1\cdot x+0\cdot y = 1$, represented by a point in parameter space as $(a,b,c)=(1,0,1)$.

The tangent to your circle at $(x,y) = (4,2)$ has $x=4$ everywhere, so can be written as a point in parameter space as $(1,0,4)$.

For more detail, see here: http://www.mathsisfun.com/algebra/line-equation-general-form.html

Emily
  • 35,688
  • 6
  • 93
  • 141
0

Let y = mx + b be the tangent line to the circle at (4,2), then : (x - 3)^2 + (mx + b - 2)^2 = 1 , and 2 = 4m + b ==> b - 2 = -4m, and : (x - 3)^2 + (mx - 4m)^2 = 1 ==> x^2 - 6x + 9 + m^2*x^2 - 8m^2*x + 16m^2 = 1 has unique solution ==> (1 + m^2)x^2 - (6 + 8m^2)x + 8 + 16m^2 = 0 ==> delta' = 0 ==> (3 + 4m^2)^2 - (1 + m^2)(8 + 16m^2) = 0. But delta' = 9 + 24m^2 + 16m^4 - 8 - 16m^2 - 8m^2 - 16m^4 = 1 > 0. this shows that for any real m, the line y = mx + 2 - 4m always intersects the circle at 2 distinct points and therefore it can not be a tangent line to the circle. The remaining case to check is the vertical line x = 4 is the tangent line. This line x = 4 goes through the point (4,2) which is on the circle. To see that it is the tangent line we see that the distance from the center of the circle to this line is 4 - 3 = 1 = the radius of the circle. So it must be the tangent line.

DeepSea
  • 77,651
-1

Since when the line x=2 started having zero gradient, that too obviously! Answer is y=1 and y=3. It can be seen very easily from graph. Mathematically any line with 0 slope is y=constant. So put y=k in equation and use condition of tangency.

Casteels
  • 11,292
  • 4
  • 27
  • 38
user131714
  • 1