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...

What is the product of the coordinates (h,k) of the center of the circle

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.

Asked on by mabottle

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

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