# Thread: Circle/linear intersection proof

1. ## Circle/linear intersection proof

Hi all,

Prove that, for any value of c, the circle whose equation is $(x+c)^2+(y-c)^2=c^2$ touches both the x-axis and the y-axis.

Is it enough here to show that $b^2-4ac=0$ when either linear equation $x=0, y=0$ is plugged into the the circle equation.

discriminant in both cases is $4c^2-4c^2=0$

or is this incorrect?

thanks sammy

2. The Equation circle centered at (m, n) wibth radius R is :

$(x-m)^2+(y-n)^2 = R^2$

Since comparative data circuit is :

$(x+c)^2+(y-c)^2 = c^2$

After all, this is a circle centered at the point (c, c-) with radius | c |.

I think that this trick question ... To know that the radius must always be positive that this is a definition (c tells us that real, not a given that the charges). For c = 0 of course it is a circle centered at the origin and radius 0, ie a point so it meets the conditions are empty. Any c different from 0 (positive or negative) point (c, c-) of course exists, then there is only that | c | then there is always such a positive cycle

3. Hello, sammy28!

Prove that, for any value of $c$, the circle $(x+c)^2+(y-c)^2\:=\:c^2$
touches both the $x$-axis and the $y$-axis.

Let $y = 0\!:\;\;(x+c)^2 + (-c)^2 \:=\:c^2 \quad\Rightarrow\quad (x+c)^2 \:=\:0$

. . . . . . . . . . $x+c \:=\:0 \quad\Rightarrow\quad x \:=\:-c$

The circle has one $x$-intercept; it is tangent to the $x$-axis.

Let $x = 0\!:\;\;(c)^2 + (y-c)^2 \:=\:c^2 \quad\Rightarrow\quad (y-c)^2 \:=\:0$

. . . . . . . . . . $y-c \:=\:0 \quad\Rightarrow\quad y \:=\:c$

The circle has one $y$-intercept; it is tangent to the $y$-axis.

Edit: I corrected my typos.
.

4. Originally Posted by Soroban

Let $x = 0\!:\;\;(c)^2 + (x-c)^2 \:=\:c^2 \quad\Rightarrow\quad (x-c)^2 \:=\:0$

. . . . . . . . . . $x-c \:=\:0 \quad\Rightarrow\quad x \:=\:c$

The circle has one $y$-intercept; it is tangent to the $y$-axis.
thanks yehoram and soroban . I like soroban's answer since it states the exact equations for all values of c.

i think the above should be

Let $x = 0\!:\;\;(c)^2 + (y-c)^2 \:=\:c^2 \quad\Rightarrow\quad (y-c)^2 \:=\:0$

. . . . . . . . . . $y-c \:=\:0 \quad\Rightarrow\quad y \:=\:c$

The circle has one $y$-intercept; it is tangent to the $y$-axis.