Circle Geom: Show Line Tangent & Find Point of Contact

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The discussion focuses on proving that the line 2x + 3y = 27 is tangent to the circle centered at (4,2) with a radius of √13. The equation of the circle is established as (x-4)² + (y-2)² = 13. The user attempts to substitute values from the tangent line into the circle's equation to derive a quadratic equation for x. They note that the discriminant of this quadratic will indicate whether the line intersects the circle at one point, confirming tangency. The solution requires demonstrating that the quadratic has a single root, which corresponds to the line being tangent to the circle.
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Homework Statement


Show that the line 2x+3y=27 is a tangent to the circle with centre (4,2) and radius sqrt of 13. Find the co-ordinates of the point of contact. (Without a calculator)


Homework Equations





The Attempt at a Solution



I have worked out that the equation of the circle is (x-4)^2 + (y-2)^2=13.
I have tryed to substitute in values of x and y from the equation of the tangent to try and work out the discriminant but I haven't been able to create an equation with just either Xs or Ys in that is easy to work out the disccriminant of without using a calculator. Please show me the method for both parts of the question.
 
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Since you are given the equation 2x+ 3y= 27, which is the same as y= 9- (2/3)x, at any point of intersection we must have (x- 4)^2+ (9- (2/3)x- 2)^2= (x- 4)^2+ (7- (2/3)x)^2= 13. Multiply that out and you have a quadratic equation for x. <br /> <br /> Now, there are three possibilities for a line and a circle:<br /> 1) they do not intersect at all.<br /> 2) they intersect in two different points.<br /> 3) they touch at one point.<br /> <br /> and those correspond to the three possiblities for solutions of a quadratic equation:<br /> 1) there are two complex roots.<br /> 2) there are two real roots.<br /> 3) there is a single root.<br /> <br /> In both situations, (3) is the case for a tangent line. Use the &quot;discriminant&quot; to show that this quadratic equation has a single solution.
 
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