# going in circles

• Jul 23rd 2009, 01:30 AM
s_ingram
going in circles
Hi folks,

I am trying to solve the following question:

Find the equations of the two circles which pass through the point (2,0) and have both the y-axis and the line y - 1 = 0 as tangents.

This seems like an easy question because you can do it by scale drawing. The first (small) circle can't really go anywhere else - see attachment! But assuming you had to do it by calculation.

To find the equation of a circle we need 3 points on the circumference. We have one supplied directly (2,0) then we have the general equation:

$\displaystyle x^2 + y^2 + 2fx + 2gy + c = 0 (1)$

plus the two tangents:

x = 0

y = 1

substituting (2,0) gives:

4 + 4f + c = 0

and substituting x = 0 into the general equation gives:

$\displaystyle y^2 + 2gy -4(f + 1) = 0$

a quadtratic in y. Since the tangent touches the circle we expect a repeated root i.e. $\displaystyle b^2 - 4ac = 0$

$\displaystyle 4g^2 + 4.1.4(f + 1) = 0$

$\displaystyle g^2 + 4f + 4 = 0 (4)$

then substitute y = 1 into the general equation:

$\displaystyle x^2 + 1 + 2fx + 2g -4(f + 1) = 0$

$\displaystyle x^2 + 2fx + 2g + 1 - 4(f + 1) = 0$

again setting $\displaystyle b^2 - 4ac = 0$ gives:

$\displaystyle 4f^2 + 16f - 8g + 12 = 0 (5)$

Equations (4) and (5) are correct. What I mean is that when you substitue the correct answers in, they give consistent results but how do you solve them? It seems clear that I am on the wrong track. I then realised that the equation y = -x + 1 must be a diameter to both circles so I tried substituting this into the general equation (1) to solve for the intersection points. It yielded a result for x with many terms in f and g that I couldn't reduce to anything like the correct answer.

Any help here would be appreciated!
best regards
• Jul 23rd 2009, 01:53 AM
running-gag
Hi

AS you said the line y=-x+1 is a diameter of the circles
The centers are on this line
Let C (xC,yC) one center of the possible circles
yC = -xC + 1

The distance between C and the point M(2,0) is such that CM² = (xC-2)²+yC² = 2xC² - 6 xC + 5
This distance is the same as the one between C and the y-axis, which is |xC|
Therefore : 2xC² - 6 xC + 5 = xC²
xC² - 6 xC + 5 = 0
(xC - 1)(xC - 5) = 0
xC = 1 or 5
• Jul 23rd 2009, 01:55 AM
earboth
Quote:

Originally Posted by s_ingram
Hi folks,

I am trying to solve the following question:

Find the equations of the two circles which pass through the point 2,0) and have both the y-axis and the line y-1=0 as tangents.

...

1. I used this equation of a circle:

$\displaystyle (x-a)^2+(y-b)^2=r^2$ with the center at C(a, b) and the radius r.

2. You already know 3 points of the circle: (2, 0), (0, 1-r), (a, 1)

3. Substitute the coordinates of these points into the equation of the circle. You'll get a system of equations. Solve this system for a, b and r.

$\displaystyle \left|\begin{array}{rcl}(2-a)^2+(0-b)^2&=&r^2 \\ (0-a)^2+(a+1-b)^2&=&r^2 \\ (a-a)^2 + (1-b)^2&=&r^2 \end{array}\right.$

4. After some transformations and calculations you'll get:

Circle #1: $\displaystyle (1,0); r=1$
Circle #2: $\displaystyle (5,-4); r=5$
• Jul 23rd 2009, 04:14 AM
Soroban
Hello, s_ingram!

Quote:

Find the equations of the two circles which pass through the point (2,0)
and have both the $\displaystyle y$-axis and the line $\displaystyle y\,=\, 1$ as tangents.

Code:

        |         |        R       - + - P - * o * - - - y=1     ----+---o-----:-----*----------         | * 2    :      *         |*        :        *         |        :         *        :        *       Q o - - - - o C      *         *      (x,y)      *         |         |*                *         | *              *         |  *          *         |      * * *         |

The center of the circle is $\displaystyle C(x,y)$

It passes through the point $\displaystyle P(2,0)$
It is tangent to the $\displaystyle y$-axis at $\displaystyle Q(0,y)$
It is tangent to $\displaystyle y = 1$ at $\displaystyle R(x,1)$

The radius is: .$\displaystyle CP \:=\: CQ \:=\: CR \:=\: r$

We have: .$\displaystyle x \:=\:r$

Since $\displaystyle CR \:=\:1-y \:=\:r\:=\:x$, then: .$\displaystyle y \:=\:1-x$

Since $\displaystyle CP \,=\,R$, we have: .$\displaystyle \sqrt{(x-2)^2 + (y-0)^2} \:=\:r$

. . which becomes: .$\displaystyle \sqrt{(x-2)^2 + (1-x)^2} \:=\:x$

Then: .$\displaystyle (x-2)^2 + (1-x)^2 \:=\:x^2$

. . which simplifies to: .$\displaystyle x^2 - 6x + 5 \:=\:0$

. . which factors: .$\displaystyle (x-1)(x-5) \:=\:0$

. . and has roots: .$\displaystyle x \:=\:1,5$

. . . . . . . . Then: .$\displaystyle y \:=\:0,\text{-}4$

One circle has center $\displaystyle (1,0)$ and $\displaystyle r = 1\!:\quad \boxed{(x-1)^2 + y^2 \:=\:1}$

The other has center $\displaystyle (5,\text{-}4)$ and $\displaystyle r = 5\!:\quad \boxed{(x-5)^2 + (y+4)^2 \:=\:25}$

• Jul 23rd 2009, 10:57 AM
s_ingram
Excellent! Thanks a lot guys.