# What is the product of the coordinates (h,k) of the center of the circleA 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...

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.

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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 <=>hÂ =-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