2230.31 – A Circle and a Parabola

The values of yy that satisfy both of these equations:

x2+y2−16=0x2−3y+12=0\begin{aligned} x^2 + y^2 - 16 &=& 0 \\ x^2 - 3 y + 12 &=& 0 \end{aligned}

are

  1. only 4
  2. -7 and 4
  3. 0 and 4
  4. no yy at all
  5. all real yy
Solution

Of course you can simply check the given answers. You can also graph the two equations. They yield the circle and parabola.

Or you can use algebra. From the second equation, we find that

x2=3y−12. x^2 = 3 y - 12.

This can be substituted into the first equation,

0=x2+y2−16=(3y−12)+y2−16=y2+3y−28=(y+7)(y−4), 0 = x^2 + y^2 - 16 = (3 y - 12) + y^2 - 16 = y^2 + 3 y - 28 = (y + 7) (y - 4),

which is a mere quadratic equation whose roots are y=4,−7y = 4, -7. But do these roots satisfy both equations?

 for y=4:x2=12−12=0⇝ OK, for y=−7:x2=−21−12=−33⇝ not OK.\begin{aligned} \text{ for } y = 4 &:& x^2 = 12 - 12 = 0 \leadsto \text{ OK,} \\ \text{ for } y = -7 &:& x^2 = -21 - 12 = -33 \leadsto \text{ not OK}. \end{aligned}

The answer is (a).