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Question

Write an equation for a circle that satisfies each set of conditions. Then graph the circle. center at (-4,-3), tangent to y=3

Solution

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The standard form of the equation of a circle with center (h,k)(h,k) and radius rr is given by:

(xh)2+(yk)2=r2(x-h)^2+(y-k)^2=r^2

We are given (h,k)=(4,3)(h,k)=(-4,-3). If a line is tangent to a circle, then the line is perpendicular to the radius drawn to the point of tangency. The distance between (4,3)(-4,-3) and y=3y=3 is 33=6|-3-3|=6 so r=6r=6:

(x(4))2+(y(3))2=62(x-(-4))^2+(y-(-3))^2=6^2

(x+4)2+(y+3)2=36\color{#c34632}(x+4)^2+(y+3)^2=36

To graph, locate the center first and use the radius to identify other points rr units to the left, right, above, and below the center:

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