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

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Tangents to the circle defined by the equation x^2 + y^2 = 169 are drawn at the points (5,12) and (5,-12). The equations of these tangents are 5x + 12y = 169 and 5x - 12y = 169, derived by determining the slopes of the lines perpendicular to the radii at those points. The intersection point of the two tangents can be found by solving the equations simultaneously, resulting in the coordinates (169/5, 0). This intersection point reflects the symmetry of the situation, confirming that the y-coordinate is zero. The discussion emphasizes the methods for finding tangent equations and their intersection in relation to the circle.
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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.
 
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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. :)
 
greg1313 said:
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.
 
RTCNTC said:
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 [math]\frac{12}{5}[/math]. Since the tangent to the circle at that point is perpendicular to that line is perpendicular to it, it has slope [math]-\frac{5}{12}[/math]. Since it goes through (5, 12) its equation is [math]y- 12= -\frac{5}{12}(x- 5)[/math]. We can write that as [math]12y- 144= -5x+ 25[/math] or [math]5x+ 12y= 169[/math].

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

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.
 
I will work on this when time allows.
 
"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.
 
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