Prove

**Punch** · September 25th 2010, 07:34 AM

show that the circle $x^2+y^2-2x-2y+4=0$ touches the $y$ -axis
I found the radius to be $\sqrt{1^2+1^2-4}=\sqrt{-2}$
is there a problem with this question?

**Archie Meade** · September 25th 2010, 08:00 AM

Originally Posted by Punch

show that the circle $x^2+y^2-2x-2y+4=0$ touches the $y$ -axis
I found the radius to be $\sqrt{1^2+1^2-4}=\sqrt{-2}$
is there a problem with this question?

Yes there is!

A circle is...

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

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

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

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

Your constant has to be <2 to have any radius at all for the circle centred at (1,1)

**Traveller** · September 25th 2010, 08:01 AM

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.

**Archie Meade** · September 25th 2010, 08:10 AM

Originally Posted by Traveller

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.

$y^2-2y+4=0$ has no real solution.

**Traveller** · September 25th 2010, 08:40 AM

Originally Posted by Archie Meade

$y^2-2y+4=0$ has no real solution.

Thanks for pointing that out. The question seems to be wrong.

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