Question
Choose the equation of a circle with radius 6 and center $(3,-5) .$$$\begin{array}{l}{\text { (a) }(x-3)^{2}+(y+5)^{2}=6} \\ {\text { (b) }(x+3)^{2}+(y-5)^{2}=36} \\ {\text { (c) }(x+3)^{2}+(y-5)^{2}=6} \\ {\text { (d) }(x-3)^{2}+(y+5)^{2}=36}\end{array}$$
Step 1
Step 1: The standard form of the equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the center of the circle and $r$ is the radius. Show more…
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Multiple Choice Choose the equation of a circle with radius 6 and center (3,-5) (a) $(x-3)^{2}+(y+5)^{2}=6$ (b) $(x+3)^{2}+(y-5)^{2}=36$ (c) $(x+3)^{2}+(y-5)^{2}=6$ (d) $(x-3)^{2}+(y+5)^{2}=36$
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Find the center, the radius, the diameter, the circumference, and the area of the circle represented by each equation. $\mathbf{a.} x^{2}+y^{2}=36$ b. $(x+5)^{2}+y^{2}=\frac{9}{4}$ c. $(x-3)^{2}+(y+6)^{2}=100$ $d. \frac{(x+5)^{2}}{3}+\frac{(y-2)^{2}}{3}=27$
Coordinate Geometry Extended
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