Explain WolframAlpha output

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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).