How Are Tangents to a Circle Found and Their Intersection Point Determined?

[mathdad]

Tangents are drawn to the circle x^2 + y^2 = 169 at the points (5,12) and (5,-12).

A. Find the equations of the tangents.

B. Find the coordinates of the point where these two tangents intersect.

I need steps for parts A and B above.

I know the equation x^2 + y^2 = 169 is a circle of radius 13. The two equations I must find touch the circle at the two given points.

 

[Greg]

Also, the circle is centered at the origin, so the two lines perpendicular to the tangents pass through (0, 0) and (5, $\pm$12). If a line is perpendicular to another line with slope m, its slope is -1/m.

Is that sufficient information for you to solve the problem? If you have any questions (on notation, perhaps) please ask. :)

 

[mathdad]

Also, the circle is centered at the origin, so the two lines perpendicular to the tangents pass through (0, 0) and (5, $\pm$12). If a line is perpendicular to another line with slope m, its slope is -1/m.

Is that sufficient information for you to solve the problem? If you have any questions (on notation, perhaps) please ask. :)

When time allows, I will post my complete reply.

 

[HallsofIvy]

Tangents are drawn to the circle x^2 + y^2 = 169 at the points (5,12) and (5,-12).

A. Find the equations of the tangents.

B. Find the coordinates of the point where these two tangents intersect.

I need steps for parts A and B above.

I know the equation x^2 + y^2 = 169 is a circle of radius 13. The two equations I must find touch the circle at the two given points.

The slope of the line through the origin (the center of the circle) and (5, 12) is \(\displaystyle \frac{12}{5}\). Since the tangent to the circle at that point is perpendicular to that line is perpendicular to it, it has slope \(\displaystyle -\frac{5}{12}\). Since it goes through (5, 12) its equation is \(\displaystyle y- 12= -\frac{5}{12}(x- 5)\). We can write that as \(\displaystyle 12y- 144= -5x+ 25\) or \(\displaystyle 5x+ 12y= 169\).

Similarly the slope of the line through the origin and (-5, 12) is \(\displaystyle -\frac{12}{5}\) so the slope or the tangent line is \(\displaystyle \frac{5}{12}\). The equation of the tangent line is \(\displaystyle y- 12= \frac{5}{12}(x+ 5)\) which we can write as \(\displaystyle 12y- 144= 5x+ 25\) or \(\displaystyle 12y- 5x= 169\).

To solve (b), find (x, y) that satisfy both 5x+ 12y= 169 and -5x+ 12y= 169. I suggest first adding the two equations to eliminate x.

 

[mathdad]

I will work on this when time allows.

 

[HallsofIvy]

"Tangents are drawn to the circle x^2 + y^2 = 169 at the points (5,12) and (5,-12).

A. Find the equations of the tangents.

B. Find the coordinates of the point where these two tangents intersect."

There are two ways to do (A).

Using Calculus: Differentiating with respect to x, 2x+ 2yy'= 0. AT (5, 12) that is 2(5)+ 2(12)y'= 10+ 24y'= 0 so y'= -10/24= -5/12. The line through (5, 12) with slope -5/12 is y- 12= (-5/12)(x- 5) or 12y- 144= -5x+ 25 which we can write as 5x+ 12y= 169. At (5, -12), that is 2(5)+ 2(-12)y'= 0 so y'= 5/12. The line through (5, -12) with slope 5/12 is y+ 12= (5/12)(x- 5) or 12y+ 144= 5x- 25 which we can write as 5x- 12y= 169.

Without Calculus, using the fact that a tangent to a circle, at a point on the circle, is perpendicular to the radius at that point: The radius at (5, 12) is the line through the center of the circle, (0, 0) to (5, 12). It has slope 12/5 so a line perpendicular to it, the tangent line to the circle, has slope -5/12. Now finish as before. Similarly for the tangent at (5, -12).

The equations of the two lines, 5x+ 12y= 169 and 5x- 12y= 169, added together, give 10x= 2(169) so x= 169/5. If we subtract the second equation from the first, we get 24y= 0 so y= 0. The two tangents intersect at (169/5, 0). The fact that the y coordinate is 0 should have been obvious from the symmetry of the situation.