Exercise 8.
"The general equation of a circle with centre (a, b) and radius r
"is x minus a all squared plus y minus b all squared
"is equal to r squared.
"By completing the square on the terms with x
"and then the terms with y, find the centre and radius of the circle."
So we have x squared minus 2x plus y squared minus 6y equals 1.
And from that, we want to try to create
this more general form for a circle,
and we're going to need to do that, as the question states,
by completing the square on the terms involving x
and also completing the square on the terms involving y.
So the terms involving x are x squared –2x.
So we're going to need to figure out what we need to put on the end here
to make that a perfect square,
and then the terms involving y are y squared minus 6y.
And again, we're going to need to figure out
what we need to put on the end here to make that a perfect square.
And ... on the other side of the equation, we have equals 1.
So x squared minus 2x.
We know to complete the square,
we're looking to halve and square the coefficient of x.
So half of –2 is –1, and the square of –1 is +1.
So adding +1 in here makes that a perfect square.
Now, we can't just add 1 willy-nilly into an equation.
So we can maintain the balance of this equation
by also adding 1 onto the right-hand side.
Similarly, considering the y terms, to complete the square,
we're looking to halve and square the coefficient of y.
So that is –6.
So half of –6 is –3.
And –3 squared is +9.
Again, we need to maintain the balance of the equation.
We can't simply add 9 into the equation on the left-hand side
without doing the same to the right.
So we'll also add 9 over here on the right-hand side.
So now we've created a perfect square with the x terms.
We have x squared minus 2x plus 1,
which is equivalent to x minus one all squared.
And we've also created a perfect square with the y terms.
We have y squared minus 6y plus 9,
which is equivalent to y minus 3 all squared.
And on the other side of the equation,
we're now left with a total of 11.
So now we've taken the equation in an expanded form
and rewritten it in a form that's much more useful
when we're dealing with circles.
So this equation is equivalent to x minus 1 all squared
plus y minus 3 all squared equal to 11.
And we know from the information provided in the question
that the centre of the circle is given by these two numbers in here.
So we have a centre ...
... at the point (1, 3).
And we know that the radius of the circle
is the square root of this number here,
so we know that the radius of this circle will be the square root of 11.
So we have a circle with a centre at the point (1,3)
and a radius of square root of 11.