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marta [7]
4 years ago
15

Write the equation of circle with the following characteristics: Center at (5,8) and Radius of 5

Mathematics
1 answer:
navik [9.2K]4 years ago
5 0

(
x
+
5
)
2
+
(
y
−
1
)
2
=
26
Explanation:
A circle of radius
r
, centered at
(
h
,
k
)
has the form
(
x
−
h
)
2
+
(
y
−
k
)
2
=
r
2
.
From the given information, we have
h
=
−
5
and
k
=
1
. To find
r
, we need only find the distance from the center
(
h
,
k
)
to a point on the circle. As
(
0
,
0
)
is such a point, applying the distance formula gives us
r
=
√
(
−
5
−
0
)
2
+
(
1
−
0
)
2
=
√
26
Thus the desired equation is
(
x
−
(
−
5
)
)
2
+
(
y
−
1
)
2
=
√
26
2
or, simplifying,
(
x
+
5
)
2
+
(
y
−
1
)
2
=
26
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Answer:

y=1/5x-31/5 (or y=0.2x-6.2 in decimals)

Step-by-step explanation:

x=5y-8     (original equation)

x+8=5y <u>(Addition Property of Equality)</u>

1/5x+8/5=y <u>(Division Property of Equality, </u><u>slope </u><u>of </u><u><em>original equation </em></u><u>is </u><u>1/5</u><u>)</u>

<u />

y-y1=m(x-x1)    <u>(point-slope formula)</u>

y-(-5))=1/5(x-(6)) <u>(plug in the slope that was found earlier and the point given in the question)</u>

y+5=1/5(x-6)

y+5=1/5x-6/5   (Distributive Property of Multiplication)   Note:  5 = 25/5

<u>y=1/5x-31/5</u> (Subtraction Property of Equality, and there's your answer)

<u>y=0.2x-6.2</u><em> </em>(This is the same answer, but written with decimals)

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Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview
svetlana [45]

Answer:  The required solution is

y=(-2+t)e^{-5t}.

Step-by-step explanation:   We are given to solve the following differential equation :

y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.

So, the general solution of the given equation is

y(t)=(A+Bt)e^{-5t}.

Differentiating with respect to t, we get

y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.

According to the given conditions, we have

y(0)=-2\\\\\Rightarrow A=-2

and

y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.

Thus, the required solution is

y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.

6 0
3 years ago
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