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Exercise 8.

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"The general equation of a circle
with centre (a, b) and radius r

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"is x minus a all squared
plus y minus b all squared

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"is equal to r squared.

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"By completing the square
on the terms with x

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"and then the terms with y, find
the centre and radius of the circle."

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So we have x squared minus 2x
plus y squared minus 6y equals 1.

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And from that,
we want to try to create

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this more general form
for a circle,

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and we're going to need to do that,
as the question states,

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by completing the square
on the terms involving x

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and also completing the square
on the terms involving y.

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So the terms involving x
are x squared –2x.

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So we're going to need to figure out
what we need to put on the end here

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to make that a perfect square,

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and then the terms involving y
are y squared minus 6y.

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And again,
we're going to need to figure out

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what we need to put on the end here
to make that a perfect square.

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And ... on the other side of
the equation, we have equals 1.

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So x squared minus 2x.

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We know to complete the square,

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we're looking to halve and square
the coefficient of x.

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So half of –2 is –1,
and the square of –1 is +1.

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So adding +1 in here
makes that a perfect square.

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Now, we can't just add 1 willy-nilly
into an equation.

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So we can maintain the balance
of this equation

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by also adding 1
onto the right-hand side.

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Similarly, considering the y terms,
to complete the square,

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we're looking to halve and square
the coefficient of y.

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So that is –6.

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So half of –6 is –3.

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And –3 squared is +9.

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Again, we need to maintain
the balance of the equation.

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We can't simply add 9 into
the equation on the left-hand side

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without doing the same
to the right.

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So we'll also add 9 over here
on the right-hand side.

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So now we've created a perfect square
with the x terms.

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We have x squared minus 2x plus 1,

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which is equivalent
to x minus one all squared.

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And we've also created
a perfect square with the y terms.

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We have y squared minus 6y plus 9,

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which is equivalent
to y minus 3 all squared.

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And on the other side
of the equation,

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we're now left with
a total of 11.

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So now we've taken the equation
in an expanded form

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and rewritten it in a form
that's much more useful

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when we're dealing
with circles.

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So this equation is equivalent
to x minus 1 all squared

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plus y minus 3 all squared
equal to 11.

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And we know from the information
provided in the question

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that the centre of the circle is
given by these two numbers in here.

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So we have a centre ...

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... at the point (1, 3).

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And we know that the radius
of the circle

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is the square root
of this number here,

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so we know that the radius of this
circle will be the square root of 11.

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So we have a circle
with a centre at the point (1,3)

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and a radius of square root of 11.