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
Asked
Active
Viewed 29 times
0
-
https://math.stackexchange.com/questions/2254073/the-line-y-mxc-is-a-tangent-to-x2y2-a2-if – lab bhattacharjee Mar 17 '19 at 11:37
2 Answers
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. $$(2mc)^2-4(1+m^2)(c^2-a^2)=0$$ $$4m^2c^2-4(c^2-a^2+m^2c^2-m^2a^2)=0$$ $$m^2c^2-c^2+a^2-m^2c^2+m^2a^2=0$$ $$c^2=a^2(m^2+1)$$
Peter Foreman
- 19,947
-
I can understand your method but please read the question once. It is given that $$c^{2}=a^{2}(m^{2}+1)$$ and you have to prove the other thing. But you have given the vice versa. – Equation_Charmer Mar 18 '19 at 06:48
-
Your answer is correct for the question" give the condition for which value of x from the equations given above are equal." – Equation_Charmer Mar 18 '19 at 06:49
0
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".