Assume $P$ is a point in line $x+y=m$, where $m \in \Bbb{R}$. There are two points $A,B$ in circle $$x^2+y^2 = 10$$ such that $PA$ and $PB$ are tangent lines of the above circle. If line: $x+y=m$ has no common point with the circle: $x^2+y^2 = 10$. Find the value range of $m$.

Assume the center of circle is $O$, then $OBPA$ is a square. so we have $OP = \sqrt{2}\times \sqrt{10} = 2\sqrt{5}$ and $OP = \sqrt{x_*^2+y_*^2}$ (where $(x_*, y_*)$ is a point in the line $x+y=m$) and then $$|m |=|x_* + y_*| \leq \sqrt{2(x_*^2+y_*^2)}=2\sqrt{10}$$ on the other hand , because the line has no common point with the circle. so $$\frac{|m|}{\sqrt{2}} \geq \sqrt{10}$$

then we get the answer: $$2\sqrt{5} \leq m \leq 2\sqrt{10}$$ and $$-2\sqrt{10}\leq m \leq -2\sqrt{5}$$

I think my answer is right. But I feel a little unpleasant. I don't know why. I think I not really understand this problem.

Is there some other way to solve this problem ? thanks very much.


Here is a more algebraic alternative:

If the line does not intersect the circle, then there are no common solutions between the two equations,

$$x+y = m$$ $$x^2 + y^2 = 10$$

If we manipulate the first equation, we get $y = m - x$. Substitute this into the second to get:

$$x^2 + (m - x)^2 = 10$$ $$x^2 + x^2 - 2mx + m^2 = 10$$ $$2x^2 - 2mx + m^2 - 10 = 0$$

Since there are no (real) solutions to this quadratic equation, its discriminant should be negative. In other words,

$$(-2m)^2 - 4(2)(m^2 - 10) < 0$$ $$-4m^2 + 80 < 0$$ $$m^2 > 20$$ $$|m| > \sqrt{20}$$

or if you prefer, $$2\sqrt{5} > m> -2\sqrt{5}$$

  • $\begingroup$ $m^2>20$ is $|m|>\sqrt{20}$ which result in two separated intervals. $m>\sqrt{20}$ or $m<-\sqrt{20}$ $\endgroup$ – rlartiga Apr 10 '14 at 16:12
  • $\begingroup$ Actually the OP make the same mistake $\endgroup$ – rlartiga Apr 10 '14 at 16:20
  • $\begingroup$ That seems to be the case $\endgroup$ – Yiyuan Lee Apr 10 '14 at 16:21

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