# Explain WolframAlpha output

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.

• That's just one of the coordinates of the two intersection points you got, no? Commented May 20, 2017 at 15:12
• True. But I still don't understand how I can make this assignment with proper arguments. Commented May 20, 2017 at 15:59
• You are working too hard for this problem. You might want to recall that a tangent and a radius can be perpendicular. Commented May 20, 2017 at 16:06
• But how do I use that information? I know they can, but I am not aware of how to implement it .. Commented May 20, 2017 at 16:11
• 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. Commented May 20, 2017 at 16:14

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 $$x_{1,2}=\frac{10}{1+k^2}\pm\sqrt{\left(\frac{10}{1+k^2}\right)^2-\frac{75}{1+k^2}}$$

• Thanks for your answer, but I do not see how to do that. Commented May 20, 2017 at 15:55
• ok a few minutes i will help you Commented May 20, 2017 at 15:56
• why the $-1$? i don't understand it Commented May 20, 2017 at 16:01
• Nope, I can't. Are you able to? :) Commented May 20, 2017 at 16:08
• I appreciate your effort, but I do not see how this explains it. :/ Commented May 20, 2017 at 16:25

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).

• I am aware of the distinction, but how did it find it? Commented May 20, 2017 at 15:37
• Your question wasn't asking for that... Commented May 20, 2017 at 15:42
• I wasn't aware of that but I have rectified it. Commented May 20, 2017 at 15:48

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: $$-10x+100=25$$ $$-10x+75=0$$ $$75=10x$$ $$\frac{15}{2}=x$$

• If you're allowed to use calculus, implicit differentiation can quickly derive the expression for the tangent line from the circle's equation. Commented May 20, 2017 at 20:14
• 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. :) Commented May 21, 2017 at 3:25
• 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. Commented May 21, 2017 at 3:51
• Of course. I will have that in mind next time I ask a question. Commented May 21, 2017 at 20:21