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3 years ago

A circle has its center at (0,0) and passes through the point (0,9). What is the standard equation of the circle?

Mathematics

1 answer:

Alexxandr [17]3 years ago

7 0

Answer:

x^2+y^2=9^2

Step-by-step explanation:

The standard equation of a circle is:

$(x-h)^2+(y-k)^2=r^2$

Where the center of a circle is (h,k) and r is the radius of the circle. In this case because a circle is equidistant from the center and we have a point where it passes through 9 that means that the radius is 9. However, since the standard equation states that we must write r in the form of $r^2$ this means that $r^2=9^2$ . Therefore by plugging in the values we have:

$(x-0)^2+(y-0)^2=9^2\\\\x^2+y^2=9^2$

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago

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emmainna [20.7K]

Answer:

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Step-by-step explanation:

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3 years ago

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kirza4 [7]

Answer:

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Step-by-step explanation:

Draw segments - 5 to A, A to the point of intersection of Band C and from from the point of intersection of B, C upto 2. It will form a square with each side 8 units and four triangles.

Subtracting the area of 4 four triangles from the area of the square will give you the area of the required shape.

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$= {8}^{2} - \frac{1}{2} \times 3 \times 6 - \frac{1}{2} \times 5 \times 2 - \frac{1}{2} \times 2 \times 3\\ - \frac{1}{2} \times 2 \times 5 \\ \\ = 64 - 3 \times 3 - 5 \times 1 - 1\times 3 - 1 \times 5 \\ \\ = 64 - 9 - 5 - 3 - 5 \\ \\ = 64 - 22 \\ = 42 \: {units}^{2}$

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