Quadratic equation solving

Question

How to prove that the value of x from $x^{2}+y^{2}=a^{2}$ and $y=mx+c$ are equal when $c^{2}=a^{2}(m^2+1)$?

I tried to equate the x from both variables but can't get it

Answer 1

$$x^2+(mx+c)^2=a^2$$ $$(1+m^2)x^2+(2mc)x+(c^2-a^2)=0$$ This has a single solution when the determinant of the quadratic is zero i.e. $$(2mc)^2-4(1+m^2)(c^2-a^2)=0$$ $$4m^2c^2-4(c^2-a^2+m^2c^2-m^2a^2)=0$$ $$m^2c^2-c^2+a^2-m^2c^2+m^2a^2=0$$ $$c^2=a^2(m^2+1)$$

Answer 2

The first locus is a circle and the second is a straight line. As shown in the post pointed to by @lab, the condition between the parameters expresses that the line tangents the circle.

This doesn't mean at all that "the values of $x$ are equal".