All categories

2 years ago

Find an equation of the circle that satisfies the given conditions.

1 answer:

4 0

The equation of a circle:
$(x-h)^2+(y-k)^2=r^2$
(h,k) - the coordinates of the centre
r - the radius

The midpoint of the diameter is the centre of a circle.
The coordinates of the midpoint:
$(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$
(x₁,y₁), (x₂,y₂) - the coordinates of endpoints

$P(-1,1) \\
x_1=-1 \\ y_1=1 \\ \\ Q(5,9) \\ x_2=5 \\ y_2=9 \\ \\
\frac{x_1+x_2}{2}=\frac{-1+5}{2}=\frac{4}{2}=2 \\ \frac{y_1+y_2}{2}=\frac{1+9}{2}=\frac{10}{2}=5$

The centre of the circle is (2,5).

The radius is the distance between an endpoint of the diameter and the centre.
The formula for distance:
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$(-1,1) \\ x_1=-1 \\ y_1=1 \\ \\ (2,5) \\ x_2=2 \\ y_2=5 \\ \\ d=\sqrt{(2-(-1))^2+(5-1)^2}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

The radius is 5.

$(x-2)^2+(y-5)^2=5^2 \\
\boxed{(x-2)^2+(y-5)^2=25}$

You might be interested in

Apply the distributive property to factor out the greatest common factor. 24 + 18+ =

irina [24]

Answer:

Step-by-step explanation:

there is no parenthesis so all that is left is to add the 2 numbers together

8 0

2 years ago

Read 2 more answers

Which number is represented by the number line

denpristay [2]

Answer: B

Step-by-step explanation: The number line has tally marks up to 3, so 3 has to be divided by something, meaning A and C do not work. There are 9 total tally marks circled, so the equation equals 9, leaving B as the only possible answer. Also, there are three tally marks per whole number, so each tally is 1/3, which is equal to 2/6, further proving it is B.

5 0

2 years ago

Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt

polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

$\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt$

$=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k$

The first integral is trivial since $(\tan t)'=\sec^2t$ .

The second can be done by substituting $u=t^2-1$ :

$u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C$

The third can be found by integrating by parts:

$u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t$

$\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8$

$\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C$

8 0

3 years ago

(-1 + 6) Squared + (4+5) squared

nalin [4]

(-1 + 6)^2 + (4+5)^2 = 106

3 0

3 years ago

A store sells 12-packs of ballpoint pens for $1.09 each and composition books for $2.79 each. Cindy buys m packs of ballpoint pe

Ludmilka [50]

There are two possible ways where you can express this. First, is by presenting the equation net of 5 percent which is "x=1.0355n+2.6505m". The other way to express such is by presenting it with gross amount which is "x= (1-0.05)(1.09n+2.79m)"

5 0

2 years ago

Read 2 more answers