A circle is tangent to the y axis and has aradius of three units. The center of the circle is in the third quadrant and lies on the graph y-2x=0
What is the product of the coordinates (h,k) of the center of the circle.
It is given that the circle is tangent to the y axis and has a radius of three units. This implies that the x-coordinate is 3 or -3.
The center lies in the third quadrant. So x = -3
The center also lies on y - 2x = 0
=> y + 6 = 0
=> y = -6
The center of the circle is (-3, -6).
The product of the coordinates of the center is -3*-6 = 18.
We notice that the location of the center of the circle is in the 3rd quadrant, therefore the values of it's coordinates are negative.
From enunciation we conclude that x = -3 <=>h =-3, because the radius of the circle is of 3 units.
We also know that the center of the circle is located also on the graph of the function y-2x=0.
y = 2x
If the center point belongs to the graph of y = 2x, it's coordinates verify the equation y = 2x.
k = 2h
Since h = -3
k = -6
The coordinates of the center of the given circle are C(-3 ; -6).
Now, we'll compute the product of the coordinates:
h*k = (-3)*(-6) = 18
The required equation of the circle is: (x+3)^2 + (y+6)^2 = 9