SOLUTION: find the equation of the circle in the fourth quadrant whose diameter is 8 and is tangent to both axes.

Question 974270: find the equation of the circle in the fourth quadrant whose diameter is 8 and is tangent to both axes.
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
If the diameter is , the radius must be .
The radii at the point of tangency with both axes, and the axes form a quadrilateral.
The radius at the points of tangency with each axis is perpendicular to that axis, so the quadrilateral has two right angles at the points of tangency.
Since the axes are perpendicular to each other,
the quadrilateral has a third right angle at the origin.
The fourth angle (at the center of the circle) has to also be a right angle,
because the measures of the interior angles of a quadrilateral add up to .
So that quadrilateral is a rectangle,
but since those radii at the points of tangency are
adjacent sides of the rectangle, and each one has a length of ,
that rectangle is a square,
and the center of the circle is at (4,-4).
The equation is <-->
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