# A circle is tangent to the $y$-axis at $y=3$ and has one $x$-intercept at $x=1$. Find the other $x$-intercept

A circle is tangent to the $y$-axis at $y=3$ and has one $x$-intercept at $x=1$. Find the other $x$-intercept

Like previously mentioned, I'm not all too familiar with circles. So, I plotted the two points and I do not know the next step. I would guess to find the slope which is $-\dfrac{3}{1}$. But that doesn't seem right. I'm just taking a shot in the dark. If someone can tell me what to do next, or at least how to find the center that would be helpful. Please do not give me the answer.

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This is a really nice problem combining high-school geometry and basic algebra, should be given to all high-school students. –  Lubin Jul 21 '12 at 1:01

Let $(a,b)$ be the centre of the circle. Since the circle is tangent to the $y$-axis at $y=3$, we must have $b=3$; can you see why? Since the known $x$-intercept is positive, the circle must lie to the right of the $y$-axis, so $a$ is positive. Thus, $a$ is just the distance from the centre of the circle to the point $(0,3)$ on the circle, so $a$ is the radius of the circle. This means that the distance from the centre at $(a,b)$ to the $x$-intercept $(1,0)$ must be $a$. This information is enough to let you solve for $a$, and once you have that, it’s not hard to find the other $x$-intercept.

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That was quick :-) –  joriki Jul 20 '12 at 21:49
Not that you really need the votes for more rep =p –  Code-Guru Jul 20 '12 at 21:55
@Brian M. Scott is a genius! That's why. –  Austin Broussard Jul 20 '12 at 21:55
@Austin: No, just experienced. And I enjoy explaining things. –  Brian M. Scott Jul 20 '12 at 22:00
I haven't been on in a few days due to a few busy days. But I am back and still am in the process of solving this problem. So is all of the circle in the first quadrant? Which would make the center of the point in the first quadrant. –  Austin Broussard Jul 24 '12 at 3:52

If (a,b) be the centre of the circle and radius=r and clearly the circle passes through (1,0)

then $r^2=(1-a)^2+b^2$

The equation of the circle $(x-a)^2+(y-b)^2=(1-a)^2+b^2$

The gradient of the circle at (x,y) = $\frac{dy}{dx} = \frac{(a-x)}{(y-b)}$

The gradient of the circle at (0,3) =$\frac{(a-0)}{(3-b)}$

As y-axis is tangent to the circle at (0,3) and its gradient is ∞, so b=3.

(i)The equation of the circle becomes $(x-a)^2+(y-3)^2=(1-a)^2+3^2$. As the circle passes through (0,3), $(0-a)^2+(3-3)^2=(1-a)^2+3^2$ =>a=5.

Or (ii) As the circle passes through (0,3), $r^2=(0-a)^2+(3-b)^2$, but $r^2=(1-a)^2+b^2$ =>a=3b-4 =>a=5

As any intersection of x-axis & the circle, (x,0)

=>$(x-a)^2+(0-b)^2=(1-a)^2+b^2$

=>x=1,2a-1.

So, the other x-intercept is 2(5)-1=9

Alternatively, the circle passes through (0,3), (1,0).

Let the equation of the circle : $x^2+y^2+2gx+2fy+c=0$

9+6f+c=0 and 1+2g+c=0.

Let the expected x-intercept be t, so the third point on the circle (t,0). So, $t^2+2gt+c=0$

So, t,1 are the roots of $s^2+2gs+c=0$

=>t+1=-2g and t.1=c

=>2g=-t-1 and c=t

As 9+6f+c=0, 2f=-$\frac{9+c}{3}$

So, the equation of the circle becomes $x^2+y^2-(t+1)x-\frac{9+t}{3}y+t=0$

The gradient of the circle at (x,y) = $\frac{dy}{dx} = \frac{3(2x-t-1)}{(9+t-6y)}$

The gradient of the circle at (0,3) =-$\frac{t+1}{t-9}$

As y-axis is tangent to the circle at (0,3) and its gradient is ∞, so t=9.

So, the other x-intercept is 9

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The question quite clearly says "Please do not give me the answer." –  Peter Phipps Jul 21 '12 at 14:51
Where should have I stopped? –  lab bhattacharjee Jul 22 '12 at 15:15
I also have not taken calculus yet or pre-calculus, so I know nothing of gradients. –  Austin Broussard Jul 24 '12 at 3:40