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