[ASK] Equation of a Circle in the First Quadrant

Apr 2018
182
38
Sukoharjo
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
Oct 2009
1,343
620
Brisbane
I'd avoid your method. Have you drawn a diagram?
 
Apr 2018
182
38
Sukoharjo
I tried drawing the diagram on paper, but it didn't help at all.
 

Debsta

MHF Helper
Oct 2009
1,343
620
Brisbane
What does "touches" the y-axis at (0, 6) mean to you?
 
Last edited:
May 2019
50
1
Mumbai (Bombay),Maharashtra State,India
Answer to your question is e.
 

Debsta

MHF Helper
Oct 2009
1,343
620
Brisbane
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
Oct 2009
1,343
620
Brisbane
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.
 
  • Like
Reactions: topsquark
Apr 2018
182
38
Sukoharjo
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?