Answer:
The center of the circle is (2/3, 0) and radius is ±6units
Step-by-step explanation:
Given the equation of a circle given as shown;
(x-2/3)²+y² = 36
To get the radius and centre of the circle, we will compare the equation with the general equation of a circle given as;
(x-a)²+(y-b)² = r² where (a, b) is the centre of the circle and r is the radius.
Comparing with the given equation,
we will see that;
a = 2/3, b = 0 and r² = 36
If r² = 36
r = ±√36
r = ±6
This shows that the center of the circle (a, b) = (2/3, 0) and radius is ±6
4, 4+4 is 8 which is double of 4
x
−
3
y
=
5
,
x
+
5
y
=
−
3
Add
3
y
to both sides of the equation.
x
=
5
+
3
y
x
+
5
y
=
−
3
Replace all occurrences of
x
with the solution found by solving the last equation for
x
. In this case, the value substituted is
5
+
3
y
.
x
=
5
+
3
y
(
5
+
3
y
)
+
5
y
=
−
3
Solve for
y
in the second equation.
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x
=
5
+
3
y
y
=
−
1
Replace all occurrences of
y
with the solution found by solving the last equation for
y
. In this case, the value substituted is
−
1
.
x
=
5
+
3
(
−
1
)
y
=
−
1
Simplify
5
+
3
(
−
1
)
.
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x
=
2
y
=
−
1
The solution to the system of equations can be represented as a point.
(
2
,
−
1
)
The result can be shown in multiple forms.
Point Form:
(
2
,
−
1
)
Equation Form:
x
=
2
,
y
=
−
1