A circle of radius $1$ centered at $(0,0)$, and $y=x^2$. What is the $x$-value of their meeting point in Quadrant I?

I thought about using trig to solve this, but failed to make much progress doing that.

Any hints?

  • $\begingroup$ Do you know the equation of a circle of radius $1$? If so, it is algebra $\endgroup$ – user142198 Feb 17 '15 at 4:44
  • $\begingroup$ what, sqrt(1-x^2)? I tried that w.o getting answer desired. I may be misunderstanding the question. $\endgroup$ – Chas Feb 17 '15 at 4:45
  • $\begingroup$ No trig, you want a polynomial equation for the circle, not one involving radicals. $\endgroup$ – Lubin Feb 17 '15 at 4:46

Hint: $$x^2+y^2=1$$ $$y=x^2$$

Where do they intersect?

  • $\begingroup$ Am I reading the Q wrong: Draw a circle of radius 1 resting in the parabola y =x^2. The distance to (a, a2) equals the radius 1 when a = _. This locates the touching point. $\endgroup$ – Chas Feb 17 '15 at 5:17
  • $\begingroup$ The $(x,y)$ have to be chosen, such that both equations are equal. $x^2 + y^2 - 1 = x^2 - y$ $\endgroup$ – user142198 Feb 17 '15 at 5:35

A unit circle will have the equation $$x^2 + y^2=1$$
To find the intersection points of this with the parabola $y=x^2$ we substitute the value of $y$.
We can write this as
$$x^2 + x^4=1$$
This is a quadratic equation in terms of $x^2$. We can now write $$x^2=\frac{-1+\sqrt{5}}{2}$$ Note that a real number squared cannot be negative. $$x=\pm\sqrt{\frac{-1+\sqrt{5}}{2}}$$
The negative solution will be eliminated as it will not be in first quadrant. Hence,

  • $\begingroup$ Looks good, but I would have went the other way of subbing in, $y+y^2=1$(personal choice of course, but it looks nicer to new students) $\endgroup$ – user142198 Feb 17 '15 at 5:47
  • $\begingroup$ I went with $x$ because this way we'd get the value of $x$ straight away. But yeah, like you said, it just personal choice. $\endgroup$ – AvZ Feb 17 '15 at 5:50
  • $\begingroup$ $\color{green}{\checkmark}\quad\quad\text{(+1)}$ from me. $\endgroup$ – user142198 Feb 17 '15 at 5:53
  • $\begingroup$ Thanks. Returned the favour :D $\endgroup$ – AvZ Feb 17 '15 at 6:00
  • $\begingroup$ Looking for the answer sqrt(3)/2. Any clue? $\endgroup$ – Chas Feb 17 '15 at 6:30

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