SOLUTION: Give the equation of the circle tangent to both axes, of radius 5 and in the 1st quadrant. In the 2nd Quadrant.In the 3rd quadrant. In the 4th quadrant.

Question 985052: Give the equation of the circle tangent to both axes, of radius 5 and in the 1st quadrant. In the 2nd Quadrant.In the 3rd quadrant. In the 4th quadrant.
Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
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The distance from each axis to center of circle must equal radius for the circle to be tangent to both axes.
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In the first quadrant, center of circle is +5 from each axis, center at (5,5)
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In second quadrant, center of circle is +5 from y axis (x=5) and 5 below x axis (y=-5), center at (5,-5)
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In third quadrant, center of circle is 5 left of y axis (x=-5) and 5 below x axis (y=-5), center at (-5,-5)
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In fourth quadrant, center of circle is 5 left of y axis (x=-5), 5 above x axis (y=5), center at (-5,5)
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Standard form of circle:

where (h,k) is the center and r is the radius
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First quadrant: h=5; k=5; r=5

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Second quadrant: h=5; k=-5; r=5

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Third quadrant: h=-5; k=-5; r=5

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Fourth quadrant: h=-5; k=5; r=5

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