Find Where Two Tangent Lines Intersect on a Circle | Math Help

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SUMMARY

This discussion provides a clear method for finding the intersection of two tangent lines on a circle. By positioning the circle's center at the origin and ensuring the tangent points share the same x-coordinate, the problem simplifies significantly. The derived formula for the distance from the center to the point of tangency is given by \( d = \frac{r^2}{x} \), where \( r \) is the radius and \( x \) is the x-coordinate of the tangent points. This approach utilizes geometric principles and algebraic manipulation to arrive at the solution.

PREREQUISITES
  • Understanding of basic geometry, specifically circles and tangents.
  • Familiarity with coordinate systems and the Cartesian plane.
  • Knowledge of algebraic manipulation and the distance formula.
  • Ability to interpret mathematical equations and diagrams.
NEXT STEPS
  • Study the properties of tangent lines to circles in geometry.
  • Learn about coordinate transformations and their applications in geometry.
  • Explore the distance formula in more depth, particularly in relation to circles.
  • Investigate other geometric constructions involving circles and tangents.
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Students studying geometry, mathematics educators, and anyone seeking to understand the intersection of tangent lines with circles.

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Here is the question:

How can i find where two tangent lines intersect on a circle?

I need math help tonight. Is there a process i need to follow to find out where the lines meet?

I have posted a link there to this topic so the OP can see my work.
 
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Hello Getty,

We can greatly simplify this problem, if we orient the circle's center at the origin of our coordinate system, and rotate the circle such that the two tangent points have the same $x$-coordinate (where $0<x<r$), one point in the first quadrant, and one in the fourth quadrant.

Please refer to the following diagram:

View attachment 1164

Because $r$ and $\ell$ are perpendicular, we may state:

$$r^2+\ell^2=d^2$$

Using the distance formula, we find:

$$\ell^2=(x-d)^2+y^2$$

and from the equation of the circle, we have:

$$y^2=r^2-x^2$$

Hence, we may now write:

$$r^2+(x-d)^2+r^2-x^2=d^2$$

$$2r^2+x^2-2xd+d^2-x^2=d^2$$

$$r^2-xd=0$$

$$d=\frac{r^2}{x}$$
 

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