Find all points of intersections of the 2 circles defined by the equations
(x - 2)2 + (y - 2)2 = 4(x - 1)2 + (y - 1)2 = 4
Answers
Step-by-step explanation:
Intersection points are
(
1
,
−
1
)
and
(
4
,
2
)
and tangents are
x
+
2
y
+
1
=
0
and
2
x
+
y
=
10
Explanation:
x
2
+
y
2
−
4
x
−
2
y
=
0
an be written as
(
x
−
2
)
2
+
(
y
−
1
)
2
=
5
and hence is a circle with center at
(
2
,
1
)
and radius
√
5
.
For finding intersection point between
x
2
+
y
2
−
4
x
−
2
y
=
0
and line
y
=
x
−
2
, we can put
y
=
x
−
2
in the equation
x
2
+
y
2
−
4
x
−
2
y
=
0
. Doing this, we get
x
2
+
(
x
−
2
)
2
−
4
x
−
2
(
x
−
2
)
=
0
or
x
2
+
x
2
−
4
x
+
4
−
4
x
−
2
x
+
4
=
0
or
2
x
2
−
10
x
+
8
=
0
or
x
2
−
5
x
+
4
=
0
or
(
x
−
1
)
(
x
−
4
)
=
0
i.e.
x
=
1
or
x
=
4
and for
x
=
1
,
y
=
−
1
and for
x
=
4
,
y
=
2
Hence intersection points are
(
1
,
−
1
)
and
(
4
,
2
)
As the slope of radius line joining
(
2
,
1
)
and
(
1
,
−
1
)
is
−
1
−
1
1
−
2
=
−
2
−
1
=
2
, slope of tangent at
(
1
,
−
1
)
is
−
1
2
and equation of tangent is
y
+
1
=
−
1
2
(
x
−
1
)
i.e.
2
y
+
2
=
−
x
+
1
or
x
+
2
y
+
1
=
0
.
And as the slope of radius line joining
(
2
,
1
)
and
(
4
,
2
)
is
2
−
1
4
−
2
=
1
2
, slope of tangent at
(
4
,
2
)
is
−
1
1
2
=
−
2
and equation of tangent is
y
−
2
=
−
2
(
x
−
4
)
i.e.
y
−
2
=
−
2
x
+
8
or
2
x
+
y
=
10
.
graph{(x^2+y^2-4x-2y)(x-y-2)(x+2y+1)(2x+y-10)=0 [-2.37, 7.63, -1.58, 3.42]}
x2 - 4x + 2 + y2 - 4y + 2 = 4 : expand equation of first circle x2 - 2x + 1 + y2 - 2y + 1 = 4 : expand equation of second circle -2x - 2y - 6 = 0 : subtract the left and right terms of the above equations y = 3 - x : solve the above for y. 2x2 - 6x + 1 = 0 : substitute y by 3 - x in the first equation, expand and group like terms. (3/2 + √(7)/2 , 3/2 - √(7)/2) , (3/2 - √(7)/2 , 3/2 + √(7)/2) : solve the above for x and use y = 3 - x to find y.