# [ASK] Equation of a Circle in the First Quadrant

#### Monoxdifly

##### Well-known member
The center of circle L is located in the first quadrant and lays on the line y = 2x. If the circle L touches the Y-axis at (0,6), the equation of circle L is ....
a. $$\displaystyle x^2+y^2-3x-6y=0$$
b. $$\displaystyle x^2+y^2-12x-6y=0$$
c. $$\displaystyle x^2+y^2+6x+12y-108=0$$
d. $$\displaystyle x^2+y^2+12x+6y-72=0$$
e. $$\displaystyle x^2+y^2-6x-12y+36=0$$

Since the center (a, b) lays in the line y = 2x then b = 2a.
$$\displaystyle (x-a)^2+(y-b)^2=r^2$$
$$\displaystyle (0-a)^2+(6-b)^2=r^2$$
$$\displaystyle (-a)^2+(6-2a)^2=r^2$$
$$\displaystyle a^2+36-24a+4a^2=r^2$$
$$\displaystyle 5a^2-24a+36=r^2$$
What should I do after this?

#### skeeter

##### Well-known member
MHB Math Helper
circle center at $(x,2x)$

circle tangent to the y-axis at $(0,6) \implies x=3$

$(x-3)^2+(y-6)^2 = 3^2$

$x^2-6x+9+y^2-12y+36 =9$

$x^2+y^2-6x-12y+36=0$

#### Monoxdifly

##### Well-known member
circle center at $(x,2x)$

circle tangent to the y-axis at $(0,6) \implies x=3$
How did you get x = 3 from (0, 6)?

#### HallsofIvy

##### Well-known member
MHB Math Helper
How did you get x = 3 from (0, 6)?
The fact that the y-axis is tangent to the circle at (0, 6) means that the line y= 6 is a radius so the y coordinate of the center of the circle is 6. And since the center is at (x, 2x), y= 2x= 6 so x= 3.

(And the center of the circle "lies on the line y= 2x", not "lays".)

#### Monoxdifly

##### Well-known member
The fact that the y-axis is tangent to the circle at (0, 6) means that the line y= 6 is a radius so the y coordinate of the center of the circle is 6. And since the center is at (x, 2x), y= 2x= 6 so x= 3.

(And the center of the circle "lies on the line y= 2x", not "lays".)
Well, I admit I kinda suck at English. I usually use "lies" as "deceives" and "lays" as "is located". Thanks for the help, anyway. Your explanation is easy to understand.