Deduced some thing...don't know what to do with it The question reads thus
Find the equation of a circle that touches the x axis at the origin and also touch the line 12x+5y=60
Now,...if i consider the centre of the circle is (a,b)
==>the radius r=|b|
That's all i could deduce
I also thought equating the equations of the cricle and the line would be a good idea to get the point of intersection...but i don't see how it could help m
Could someone help me solve the question and explain the resaoning behind so that i can know what to do next time
In addition...in case we need to find the chord of intersection of two circles are we just supposed to equate them?
Thanks in advanced
Lee
 A: The general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ with center $(-g, -f)$ in this case boils down to $x^2 + y^2 - 2by = 0$ as circle passes through origin ($c=0$) and the center is $(0,b)$ (as you have apparently deduced correctly).
Now use the tangency condition of the line $12x+5y = 60$ to the circle : perpendicular distance of the center of the circle from the line will be equal to the radius of the circle.  
Hence,
$$\frac{|5b - 60|}{13} = b \implies b =  \frac{10}{3}  \ or \frac{-15}{2}$$
As to the second part of your question, yes you are right (if the two circles intersect at two points, the common chord is equivalent to the radical axis.), but the condition is that the both the circles' equation should be in the standard form (coefficient of $x^2$ and $y^2$ should be $1$). See this reference for a simple proof.
A: I like Shraddheya Shendre's solution better than what I am giong to show, but this is an alternative.
we have two equations:
$12x + 5y = 60\\
x^2 + y^2 = 2by$
If you find the intersection:
$x = 5 - \frac {5}{12} y\\
(5 - \frac {5}{12} y)^2 + y^2 = 2by$
Yields a quadratic equation.
If the line is tangent to the circle, then the discriminant must equal 0.
$(\frac {25}{144}+1)y^2 - (2b+ \frac {50}{12}) y + 25 = 0$
$(\frac {24b + 50}{12})^2 - 4(\frac {169}{144})(25) = 0\\
576 b^2 + 2400b - 14400=0\\
6b^2 +25b - 150=0$
$b = $$\frac {-25 \pm \sqrt {625 +3600}}{12}
\frac {-25 \pm 65}{12}\\
b = -\frac {15}{2} , \frac {10}{3}$
