show that the circle $\displaystyle x^2+y^2-2x-2y+4=0$ touches the $\displaystyle y$-axis

I found the radius to be $\displaystyle \sqrt{1^2+1^2-4}=\sqrt{-2}$

is there a problem with this question?

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- Sep 25th 2010, 07:34 AMPunchProve
show that the circle $\displaystyle x^2+y^2-2x-2y+4=0$ touches the $\displaystyle y$-axis

I found the radius to be $\displaystyle \sqrt{1^2+1^2-4}=\sqrt{-2}$

is there a problem with this question? - Sep 25th 2010, 08:00 AMArchie Meade
Yes there is!

A circle is...

$\displaystyle \left(x-x_c\right)^2+\left(y-y_c\right)^2=r^2$

$\displaystyle x^2-2x_cx+\left(x_c\right)^2+y^2-2y_cy+\left(y_c\right)^2=r^2$

From your equation.... $\displaystyle \left(x_c,y_c)=(1,1)$

hence $\displaystyle x^2-y^2-2x_cx-2y_cy+2-r^2=0\Rightarrow$

Your constant has to be <2 to have any radius at all for the circle centred at (1,1) - Sep 25th 2010, 08:01 AMTraveller
If the circle touches the y-axis at some point, then for that point x = 0. Substitute this value of x in the given equation and solve for y to get a solution in reals.

- Sep 25th 2010, 08:10 AMArchie Meade
- Sep 25th 2010, 08:40 AMTraveller