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=0What is the product of the coordinates (h,k) of the...

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 jane1992

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Since the center of the circle is located in the 3rd quadrant, 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 is on 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 equation of the circle is:

(x+3)^2 + (y+6)^2 = 9

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neela | High School Teacher | (Level 3) Valedictorian

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Let  (h,k) be the centre of the circle. Since y  axis is the tangent,  The normal from the tangent || to x axis passes through the centre of the circle. So h =| r| , the radius of the circle, radius is given to be 3.

Therefore h = r = |3 |.   h should be in 3rd quadrant . So h = -3.

Therefore the centre (h,k) = (-3,k) lies on y-2x. So (-3,k) should satisfy y-2x = 0.

So k -2(-3) = 0.  Or k+5 = 0 . So k =-5.

Therefore the centre is at (-3, -5) which in 3rd quadrant.

Therefore the  equation of the is (x-h)^2 +(y-k) = r^2 . Or

{x-(-3)}^2 +{y- (-5)}^2 = 3^2.

(x+3)^2+(x+5)^2 = 3^2.

x^2+6x+9+y^2+10y+25 = 9.

Rearranging the standard form, we get the equation of the circle,

x^2+y^2+6x+10y +25 = 0 which has the centre at (-3, -5) and radius  3 units, toucching y  axis.

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