Circle - finding the equation Question:
Find the equation of a circle whose center is in the first quadrant; touches the x-axis at (4,0) and makes an intercept of length 6 units on the y-axis.
I am getting a faint idea where to start on this question, but the method turns out to be very long. I'm pretty sure that there's a small hint that will make the question much easier to solve? Please help!
 A: I think I know what you're trying to solve. So you want to find the equation of a circle such that the circle is sitting tangent to the x-axis at $\left(4,0\right)$ and its two intercepts of the y-axis must result in a line of length $6$ between them. In other words, you need to know its radius so that you can find the center and therefore its equation. Well, you already have the x-value, $4$, so no need to worry about that. To find the radius you'll need to draw a triangle connecting the center of your proposed circle to both of the y-intercepts. From there you know the length of the distance between them to be $6$, and the height of that triangle to be $4$ units. Then by simply dividing $6$ in two you arrive at one of the two triangle's bases, and then the Pythagorean theorem allows you to arrive at the missing side's length, which is the radius of your circle:
$$ 3^2+4^2=r^2\implies{r}=\sqrt{25}=\boxed{5} $$
So the equation of your circle is thus
$$ 25=\left(x-4\right)^2+\left(y-5\right)^2 $$
Just drew a quick picture, I hope I'm right...double check me.
A: A start: Draw a picture. The centre of the circle is $(4,a)$ for an as yet unknown positive $a$ and the radius is $a$. So the circle has equation $\dots$. 
In the equation, set $x=0$. The two $y$-values differ by $6$. 
Remark: The above is an analytic geometry approach. There is a much cleaner way using just basic geometry.
A: Let $O$ be the center of the circle, and let $A,B$ be the intersection points between the circle and $y$-axis. Also, let $M$ be the middle point of the line segment $AB$.
Then, letting $r$ be the radius, consider a right triangle $OMA$ with $$OA=r,\ OM=4,\  AM=6/2=3.$$
You'll get $r^2=3^2+4^2$ with the center $(4,r)$.
Hence, you'll get $r=5$. So, the answer is $(x-4)^2+(y-5)^2=25$.
A: Your first two requirements basically say that if $r$ is the radius of the circle then its center is at the point $(4,r)$ (and also that $r>0$). We put this into the standard equation of a circle and get
$${(x - 4)^2} + {(y - r)^2} = {r^2} $$
The intercept requirement means that the point $(0,6)$ is on the circle: substituting we get
$${(0 - 4)^2} + {(6 - r)^2} = {r^2}$$
Solving, we get a linear equation (the $r^2$'s cancel) which solves to
$$r=\frac{{13}}{3}$$
Substituting that into our original equation,
$${(x - 4)^2} + {(y - \frac{{13}}{3})^2} = \frac{{169}}{9}$$
You can multiply both sides by 9 and do other operations if you prefer another form for the equation.
