Find the equation of a circle touching the y-axis and passing through the points (3,1) and (6,4).
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The general equation of a circle is
(
x
−
a
)
2
+
(
y
−
b
)
2
=
r
2
Where the centre is (a,b) and the radius is r
As the circle touches the y axis r=a
Draw any circle touching the y axis to see this.
So multiplying out gives:
x
2
−
2
a
x
+
a
2
+
y
2
−
2
b
y
+
b
2
=
a
2
This simplifies to
x
2
−
2
a
x
+
y
2
−
2
b
y
+
b
2
=
0
The point (1,5) is on the circle so substitute x =1 and y=5
Likewise (8,12) is on the circle.
You will have 2 simultaneous equation to find a and b.
Eliminate a to get to
56
b
=
7
b
2
So b=8 or 0
Think about b= 0
The centre of the circle is on the y axis and the circle touches the y axis!!!
So b=8 and you can calculate the value of a
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