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My goal is to find all values of "a" so that the circle $x^2 - ax + y^2 + 2y = a$ has the radius 2

The correct answer is: $a = -6$ and $a = 2$

I tried solving it by doing this:
$x^2 - ax + y^2 +2y=a$
$x^2 - ax + (y+1)^2-1=a$
$(x - \frac a2)^2 - (\frac a2)^2 + (y+1)^2-1=a$
$(x - \frac a2)^2 - {a^2\over 4} + (y+1)^2-1=a$
$(x - \frac a2)^2 + (y+1)^2=a + {a^2\over 4} + 1$
$(x - \frac a2)^2 + (y+1)^2={a^2+4a + 4\over 4}$

We want the radius to be 2 so set this ${a^2+4a + 4\over 4}$ equal to 2
${a^2+4a + 4\over 4}=2$
$a^2+4a + 4=8$
$a^2+4a -4=0$

Solve for a:
$a=-2 \pm \sqrt{4+4}$
$a=-2 \pm \sqrt{8}$

This is not correct as you can see. I don't understand what I do wrong, I'm not sure if there is one of those tiny mistakes somewhere in my solving process or if I'm completely wrong from the beginning. Thanks in advance.

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    $\begingroup$ The equation for a circle with radius $r$ is going to be something like $(x-h)^2 + (y-k)^2 = r^2$. Note the $r^2$ on the right-hand side. Check that against your own work. $\endgroup$
    – Xander Henderson
    Commented Sep 8, 2022 at 16:14

2 Answers 2

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$\frac{a^2+4a+4}{4}$ is not a radius. Actually, it is the square of radius.

So, you should solve $\frac{a^2+4a+4}{4}=2^2$

And its solution is a=-6, a=2

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$\frac{a^2+4a+4}{4}$ should equal $2^2=4$, not $2$.

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