Video Transcript
Describe the general equation of
the shown circle 𝑀 passing through the origin point and the two points 𝐴: eight,
zero and 𝐵: zero, negative 10.
So we’ve been asked to find the
equation of this circle 𝑀. And we know three points that lie
on its circumference, the origin point zero, zero and the two points 𝐴 and 𝐵 whose
coordinates have been given. In order to find the general
equation of this circle, we’ll first find it in center–radius form. For which we need to know the
center of the circle that has the coordinates of the point 𝑀 and also the
radius. That would be the length of any of
the lines 𝑀𝐴, 𝑀𝑂, or 𝑀𝐵.
We’ll recall, first of all, that
the perpendicular bisector of any chord passes through the center of the circle. The lines 𝑂𝐴 and 𝑂𝐵 are each
chords of this circle. So if we can find the perpendicular
bisectors of these two lines, they’ll meet in the center of the circle, the point
𝑀. Let’s consider the perpendicular
bisector of the line 𝑂𝐴 first of all. Remember, perpendicular bisector is
a line which divides another line exactly in half. That’s what we mean by bisect. And the perpendicular part means
that it meets the line at right angles.
The line 𝑂𝐴 is just a horizontal
line. It lies along the 𝑥-axis. It connects the origin, the point
zero, zero, with 𝐴, the point eight, zero. As such, its perpendicular bisector
will be a vertical line passing through the midpoint of 𝑂𝐴. The 𝑥-coordinate of the midpoint
of 𝑂𝐴 will just be the average of the 𝑥-coordinates of the origin and 𝐴. That’s the average of zero and
eight, which is four. The 𝑦-coordinate will be zero as
this point is on the 𝑥-axis. As the perpendicular bisector of
𝑂𝐴 is a vertical line passing through four on the 𝑥-axis, its equation is
therefore 𝑥 equals four.
We can apply similar logic to work
out the equation of the perpendicular bisector of 𝑂𝐵. 𝐵 is the point with coordinates
zero, negative 10, which means the line 𝑂𝐵 is a vertical line. It lies along the 𝑦-axis. The perpendicular bisector of 𝑂𝐵
will be a horizontal line passing through the midpoint of 𝑂𝐵. The 𝑥-coordinate of this point
will be zero as we’re on the 𝑦-axis. And the 𝑦-coordinate will be the
average of the 𝑦-coordinates of the origin and the point 𝐵. That’s the average of zero and
negative 10, which is negative five.
Therefore, the equation of the
perpendicular bisector of 𝑂𝐵 is 𝑦 equals negative five because it is a horizontal
line passing through negative five on the 𝑦-axis. The equations of these two lines
also give the coordinates of the point 𝑀 as, remember, the point 𝑀 is on both
lines. The 𝑥-coordinate of 𝑀 will be
four. And the 𝑦-coordinate of 𝑀 will be
negative five. So we now know that the center of
our circle is the point with coordinates four, negative five.
But what about the radius? Well, the radius is the distance
from the center of the circle to any point on the circumference. So, as we said before, we could
find the length of either 𝑀𝐴, 𝑀𝑂, or 𝑀𝐵. It will be most straightforward to
work out the length of 𝑀𝑂 as the coordinates of the point 𝑂 are zero, zero. To find the distance 𝑑 between two
points on a coordinate grid, we use an application of the Pythagorean theorem. If the two points have coordinates
𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two, then the distance squared is equal to 𝑥 two
minus 𝑥 one squared plus 𝑦 two minus 𝑦 one squared.
So, using the points four, negative
five and zero, zero, that’s 𝑀 and the origin, we have that the radius squared is
equal to four minus zero squared plus negative five minus zero squared. That’s just equal to four squared
plus negative five squared. Four squared is 16. And negative five squared is
25. Remember, a negative multiplied by
another negative gives a positive. 16 plus 25 is 41. So we have that the radius of our
circle squared is equal to 41. Now, we could square root both
sides of this equation to find out what the radius of the circle is equal to rather
than the radius squared. But we actually don’t need to
because, in the next stage of our working, we’re going to be using the radius
squared.
Next, we’ll recall the
center–radius form of the equation of a circle. If a circle has its center at the
point with coordinates ℎ, 𝑘 and a radius of 𝑟 units, then its equation in
center–radius form is given by 𝑥 minus ℎ all squared plus 𝑦 minus 𝑘 all squared
equals 𝑟 squared. We know the values of ℎ and 𝑘. Remember, our circle has a center
at the point four, negative five. So ℎ is four. And 𝑘 is negative five. We’ve also found that 𝑟 squared is
equal to 41. So we can substitute each of these
values into the center–radius form of the equation of a circle.
I’m going to delete some of the
working on the screen to make space to do this. But make sure you’ve got it all
written down. So we substitute four for ℎ,
negative five for 𝑘, and 41 for 𝑟 squared. And we have that the equation of
the circle is 𝑥 minus four squared plus 𝑦 minus negative five squared equals
41. In the second bracket, 𝑦 minus
negative five is the same as 𝑦 plus five. So our equation becomes 𝑥 minus
four squared plus 𝑦 plus five squared equals 41.
This is the equation of the circle
in center–radius form. But we’ve been asked for the
general equation, which means we need to expand the brackets and simplify the
result. Let’s think about expanding these
brackets. Remember, 𝑥 minus four all squared
means 𝑥 minus four multiplied by 𝑥 minus four. We must make sure we multiply each
term in the first bracket by each term in the second. We don’t just square each term. We can use a method such as the
FOIL method to help with our expansion.
First, we multiply the first terms
in the two brackets together. That’s 𝑥 multiplied by 𝑥 which
gives 𝑥 squared. Then, we multiply the outer terms
together. That’s the 𝑥 in the first bracket
and the negative four in the second, giving negative four 𝑥. Then, we multiply the inner terms
together. That’s the negative four in the
first bracket and the 𝑥 in the second, giving another lot of negative four 𝑥. And, finally, we multiply the last
terms in the brackets together. Negative four multiplied by
negative four is 16.
We can simplify by grouping
together the like terms in the center of our expansion. Negative four 𝑥 minus four 𝑥 is
equal to negative eight 𝑥. So we have that 𝑥 minus four all
squared is equal to 𝑥 squared minus eight 𝑥 plus 16. You can use the same method to show
that the expansion of 𝑦 plus five squared is equal to 𝑦 squared plus 10𝑦 plus
25. So substituting the expansions for
the two brackets gives 𝑥 squared minus eight 𝑥 plus 16 plus 𝑦 squared plus 10𝑦
plus 25 equals 41.
Now, we can do some simplification
because on the left-hand side we have plus 16 plus 25 which is equal to 41. This will cancel with the positive
41 on the right-hand side, which we can also see by subtracting 41 from each side of
the equation. When we’re given the equation of a
circle in its general form, it’s usual to write the terms in a particular order. We start with the 𝑥 squared terms,
then the 𝑦 squared, then the 𝑥s, then the 𝑦s, and then the constant term if there
is one.
So, reordering the terms into this
standard order gives the equation of our circle as 𝑥 squared plus 𝑦 squared minus
eight 𝑥 plus 10𝑦 is equal to zero. Remember, we have no constant term
because the positive 16 and positive 25 on the left-hand side combine to cancel with
the 41 on the right-hand side. So, we’ve found that the general
equation of the circle shown, which passes through the origin and the points eight,
zero and zero, negative 10, is 𝑥 squared plus 𝑦 squared minus eight 𝑥 plus 10𝑦
equals zero.
