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I just inputted (x-10)^2+(y-2)^2 = 5^2, y= k*x+2 into the WolframAlpha input text field. Under the section dedicated to Real solutions it states that $x=\frac{15}{2}$. I do not understand this output, although I do recognise and agree with it. Can someone please help me understand the output (as in how it found it)? Thanks in advance.

Edit:

The assignment is

A circle has the equation $(x-10)^2+(y-2)^2=5^2$. Line $t$ goes through the point $p(0,2)$ and is tangent to the circle.

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  • $\begingroup$ That's just one of the coordinates of the two intersection points you got, no? $\endgroup$ Commented May 20, 2017 at 15:12
  • $\begingroup$ True. But I still don't understand how I can make this assignment with proper arguments. $\endgroup$
    – Pyrros
    Commented May 20, 2017 at 15:59
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    $\begingroup$ You are working too hard for this problem. You might want to recall that a tangent and a radius can be perpendicular. $\endgroup$ Commented May 20, 2017 at 16:06
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    $\begingroup$ But how do I use that information? I know they can, but I am not aware of how to implement it .. $\endgroup$
    – Pyrros
    Commented May 20, 2017 at 16:11
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    $\begingroup$ You've already represented the tangent as $y=kx+2$, no? Then the radius would be (why?) $y-2=-\frac1{k}(x-10)$. The intersection of these two lines is the tangency point on the circle, which should be in terms of $k$. Plug into your circle equation, and solve. $\endgroup$ Commented May 20, 2017 at 16:14

3 Answers 3

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HINT: solve the equation $$(x-10)^2+(kx+2-2)^2=25$$ for $x$ simplifying you get the following quadratic equation $$x^2(1+k^2)-20x+75=0$$ can you solve this? after dividing by $$1+k^2\ne 0$$ we get $$x^2-\frac{20x}{1+k^2}+\frac{75}{1+k^2}=0$$ we get $$x_{1,2}=\frac{10}{1+k^2}\pm\sqrt{\left(\frac{10}{1+k^2}\right)^2-\frac{75}{1+k^2}}$$

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  • $\begingroup$ Thanks for your answer, but I do not see how to do that. $\endgroup$
    – Pyrros
    Commented May 20, 2017 at 15:55
  • $\begingroup$ ok a few minutes i will help you $\endgroup$ Commented May 20, 2017 at 15:56
  • $\begingroup$ why the $-1$? i don't understand it $\endgroup$ Commented May 20, 2017 at 16:01
  • $\begingroup$ Nope, I can't. Are you able to? :) $\endgroup$
    – Pyrros
    Commented May 20, 2017 at 16:08
  • $\begingroup$ I appreciate your effort, but I do not see how this explains it. :/ $\endgroup$
    – Pyrros
    Commented May 20, 2017 at 16:25
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You are asking wolfram for which values of $x,y$ and $k$ the equations $(x-10)^2+(y-2)^2 = 5^2$ and $y= kx+2$ are true. These are 2 equations in 3 different variables so there are infinitely many solutions. Under real solutions, wolfram shows you the values of $x,y$ $k$ such that the equations are true and such that $x,y$ and $k$ are real numbers (no complex part, see complex numbers).

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  • $\begingroup$ I am aware of the distinction, but how did it find it? $\endgroup$
    – Pyrros
    Commented May 20, 2017 at 15:37
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    $\begingroup$ Your question wasn't asking for that... $\endgroup$ Commented May 20, 2017 at 15:42
  • $\begingroup$ I wasn't aware of that but I have rectified it. $\endgroup$
    – Pyrros
    Commented May 20, 2017 at 15:48
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I just managed to find out why: Tangents to the circle $(x-10)^2+(y-2)^2=5^2$ have the equation $$(x-10)(x_P-10)+(y-2)(y_P-2)=5^2$$ I can insert values for $y_P$ and $x_P$: $$(x-10)(0-10)+(y-2)(2-2)=5^2$$ where the right side equals $0$. Then I isolate for x: $$-10x+100=25$$ $$-10x+75=0$$ $$75=10x$$ $$\frac{15}{2}=x$$

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  • $\begingroup$ If you're allowed to use calculus, implicit differentiation can quickly derive the expression for the tangent line from the circle's equation. $\endgroup$ Commented May 20, 2017 at 20:14
  • $\begingroup$ Yea, unfortunately, I have not yet learned about differentiation, calculus or other higher level stuff. I guess that would also be smart to tag? ;) I actually just about started Cartesian geometry (in the plane). However, I greatly appreciate your sharing of knowledge. :) $\endgroup$
    – Pyrros
    Commented May 21, 2017 at 3:25
  • $\begingroup$ That's why supplying context is important when asking a question on this site. When I gave you the hint above in the comments, I didn't know that you knew the compact formula for a circle tangent, so I gave you the usual "blind" method to proceed. In any case, you should still try finishing that route, even if only for practice in algebraic manipulation. $\endgroup$ Commented May 21, 2017 at 3:51
  • $\begingroup$ Of course. I will have that in mind next time I ask a question. $\endgroup$
    – Pyrros
    Commented May 21, 2017 at 20:21

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