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ozzi
3 years ago
14

Can someone please help me, I don’t know how to do it, and my test is due today

Mathematics
1 answer:
tresset_1 [31]3 years ago
4 0

Answer:

what's the question? :/

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Find the equation of a circle that has a diameter with the endpoints given by the points A(-4,9) and B (- 2, - 3) )
Aleksandr-060686 [28]

The equation of the circle is (x+3)^{2}+(y-3)^{2}=37

Explanation:

Given that the endpoints of the circle are A(-4,9) and B(-2,-3)

We need to determine the equation of the circle.

<u>Center:</u>

The center of the circle can be determined using the midpoint formula,

Center=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})

Substituting the coordinates A(-4,9) and B(-2,-3), we get,

Center=(\frac{-4-2}{2},\frac{9-3}{2})

Center=(\frac{-6}{2},\frac{6}{2})

Center=(-3,3)

Thus, the center of the circle is (-3,3)

<u>Radius:</u>

The radius of the circle can be determined using the distance formula,

r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}

Substituting the center (-3,3) and the endpoint (-4,9), we get,

r=\sqrt{\left(-4+3\right)^{2}+\left(9-3\right)^{2}}

r=\sqrt{\left(-1\right)^{2}+\left(6\right)^{2}}

r=\sqrt{1+36}

r=\sqrt{37}

Thus, the radius of the circle is \sqrt{37}

<u>Equation of the circle:</u>

The standard form of the equation of the circle is given by

(x-a)^{2}+(y-b)^{2}=r^{2}

where (a,b) is the center and r is the radius.

Substituting the values, we have,

(x+3)^{2}+(y-3)^{2}=(\sqrt{37})^{2}

(x+3)^{2}+(y-3)^{2}=37

Thus, the equation of the circle is (x+3)^{2}+(y-3)^{2}=37

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3 years ago
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Can someone please help me? I keep losing points...
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Answer:

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Step-by-step explanation:

i=prt

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(40,000)(0.09)(4)

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