solutions for equations for circles

Question

I've been going through a textbook's questions about equations of circles which take the form of ${x^2 + y^2 = r^2}$

I am asked for each question to supply 2 values for each letter.

The two questions I am stuck on are:

${(t, 2t)}$ lies on ${x^2 + y^2 = 5}$

When I plug this into the circle equation, I get

${t^2 + 2t^2 = 5^2}$

${3t^2 = 5^2}$

The answer in the textbook is ${\pm 1}$ but I don't understand how that answer was obtained.

The next question is ${(p + 1, p - 1)}$ lies on ${x^2 + y^2 = 6}$

So if I plug the x and y values into the equation I get:

${(p + 1)^2 + (p - 1)^2 = 6}$

If I expand this out I get:

${p^2 + 2p + 1 + p^2 + 1 = 6}$

=> ${2p^2 + 2p + 2 = 6}$

I divide by 2 to get:

${p^2 + p + 1 = 3}$

=> ${p^2 + p -2 = 0}$

If I complete the square I get

=> ${(p + 2)(p -1) = 0}$

=> ${p = 2}$ or ${p = -1}$

But the textbook gives the answer ${\pm \sqrt 2}$

Can anyone tell me where I have gone wrong?

Answer 1

If $x^2 + y^2 = 5$ and $(x,y)$ lie on the point $(t,2t)$, then you need to replace $y$ with $2t$, so you get $$t^2 + (2t)^2 = 5$$ which is slightly different than what you got.

For the second, $$(p+1)^2 + (p-1)^2 = 6$$ expands differently than you expanded it:

• $(p+1)^2 = p^2 + 2p + 1$
• $(p-1)^2 = p^2 - 2p + 1$
• Therefore, $(p+1)^2 + (p-1)^2$ is equal to...

Answer 2

${(t, 2t)}$ lies on ${x^2 + y^2 = 5}$

then ${t^2 + (2t)^2 = 5}$

${5t^2 = 5^2}$ thus $t^2=\frac{5}{5}=1$ so $t=‎\pm{1}$

${(p + 1)^2 + (p - 1)^2 = 6}$

then

${p^2 + 2p + 1 + p^2 + 1-2p = 6}$

=> ${2p^2 + 2 = 6}‎\Rightarrow‎‎‎‎ 2p^2 = 6-2=4‎\Rightarrow‎‎‎‎ p^2=\frac{4}{2}=2 ‎\Rightarrow‎‎‎‎ p=‎\pm\sqrt{2}$