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[ASK] Equation of a Circle in the First Quadrant

Monoxdifly

Well-known member
Aug 6, 2015
284
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
Mar 1, 2012
696
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
Aug 6, 2015
284

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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
Aug 6, 2015
284
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.