#### Monoxdifly

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?

#### Debsta

MHF Helper
I'd avoid your method. Have you drawn a diagram?

#### Monoxdifly

I tried drawing the diagram on paper, but it didn't help at all.

#### Debsta

MHF Helper
What does "touches" the y-axis at (0, 6) mean to you?

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MHF Helper

#### Monoxdifly

What does "touches" the y-axis at (0, 6) mean to you?
It intersects the Y-axis only at (0, 6).

The steps?

Ummm... Can you tell me how to arrive to that circle equation without drawing first?

#### Debsta

MHF Helper
It intersects the Y-axis only at (0, 6).
Yes, in other words the y axis is a tangent to the circle, yes?

What do you know about the tangent to a circle and the radius drawn from the point of contact to the centre of the circle?

#### Debsta

MHF Helper
Ummm... Can you tell me how to arrive to that circle equation without drawing first?
No! Why would you want to do it that way when, using a diagram and a little bit of knowledge about the property of circles, you can arrive at the answer in a few simple steps.

topsquark

#### Monoxdifly

What do you know about the tangent to a circle and the radius drawn from the point of contact to the centre of the circle?
They are perpendicular, so we need to make a line perpendicular to Y-axis from (0, 6) which crosses the line y = 2x. Then?