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2 years ago

A circle has its center at (-1, 2) and a radius of 3 units. What is the equation of the circle?

Mathematics

2 answers:

Aleks [24]2 years ago

8 0

Hello,

Answer D

(x+1)²+(y-2)²=9

Irina18 [472]2 years ago

5 0

Answer:

$(x+1)^2+(y-2)^2=9$

last option is correct.

Step-by-step explanation:

We have been given that

Center: (-1,2)

Radius : 3 units

We know that the standard form of the circle is given by

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

Here, h = -2, k = 2, and r=3

Substituting these values in the above formula, we get

$(x+1)^2+(y-2)^2=3^2\\\\(x+1)^2+(y-2)^2=9$

Therefore, last option is correct.

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3 years ago

Eric can mow the Johnson’s lawn in 2 hours. It takes him 5 hours to mow the Smith’s lawn. If he is paid $9 per hour , how much w

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3 years ago

Match each trig function to a ratio. tan(A)=

ehidna [41]

Answer:

$\frac{y}{x}$ , the fourth option.

Step-by-step explanation:

Tan means $\frac{opposite}{adjacent}$ or opposite side divided by adjacent side.

Tan A equals the opposite side of the angle A divided by its adjacent side.

The side opposite of A is y and side adjacent to it is x.

So, Tan A equals $\frac{y}{x}$ !

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2 years ago

Find all points on the x-axis that are 14 units from the point (6,-7) All points on the x-axis that are 14 units from the point

Maksim231197 [3]

Answer: $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

or (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

$\sqrt{(x-6)^2+(0-(-7))^2}=14\ \ \ [\ \because distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]$

Taking square on both the sides , we get

$(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0$

Using quadratic formula : $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}$

so, $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

since $\sqrt{3}=1.732$

so, $(6+7(1.732),0)\text{ and }(6-7(1.732),0)$ are the required points.

i.e. (18.124,0) and ( -6.124,0) are the required points.

3 0

3 years ago