### Video Transcript

Find the equation of the circle that passes through the points π΄: negative four, negative three; π΅: negative three, negative four; and πΆ: negative two, negative three.

So we have the coordinates of three points on the circumference of a circle and weβre looking to find its equation. In order to do so, we need to recall a key circle theorem. Suppose you have two points on the circumference of a circle. If you join them together, you create a chord. if you then construct the perpendicular bisector of a chord, so that is the line that is both perpendicular to it at right angles and cutting it directly in half, then this line will always pass through the centre of the circle. However, we know three points on the circumference of the circle. So our approach is going to be this: weβll look at two of the chords π΄π΅, π΅πΆ and π΄πΆ, weβll find the equations of two of the perpendicular bisectors, weβll then solve them simultaneously to find the coordinates of the point where they meet, which will give us the coordinates of the centre of the circle.

So I only need to use two of the chords, so Iβm gonna start with π΄π΅. So the perpendicular bisector, well because it bisects the chord, it passes through the midpoint. So we need that first of all. So the midpoint of π΄π΅ is negative four plus negative three over two, negative three plus negative four over two. And so this gives a midpoint with coordinates negative seven over two, negative seven over two. The slope is change in π¦ over change in π₯, so subtracting the coordinates of π΄ from those of π΅, I have negative four minus negative three over negative three minus negative four which is negative one over one and therefore simplifies to just negative one. Now thatβs the slope of the chord π΄π΅, and I want the slope of the perpendicular bisector. So the slope of a perpendicular line will be the negative reciprocal of this value, so itβs just one.

So hereβs what we know about the perpendicular bisector of π΄π΅. It has a slope of one and it passes through the point with coordinates negative seven over two, negative seven over two. So if we recall the slope-intercept form of the equation of a line, it would be π¦ equals ππ₯ plus π. And here, π, our slope, is one. So the equation of our line is π¦ equals π₯ plus π.

Now looking at the coordinates of the midpoint, which is a point on this line, we see the π₯ and π¦ are both the same; theyβre both negative seven over two, which means that the value of this constant, π, is equal to zero as π¦ and π₯ equal to each other. So we have the equation of the perpendicular bisector of π΄π΅, is just π¦ equals π₯.

Next, letβs look at finding the perpendicular bisector of the chord π΄πΆ. Now these two points actually have the same π¦-coordinate of negative three, which means that the line joining these two points is a horizontal line. Therefore, the perpendicular bisector of this line will be a vertical line passing through the midpoint. The midpoint of π΄πΆ has coordinates negative four plus negative two over two, negative three plus negative three over two which is the point with coordinates negative three, negative three. Now weβve already discussed that the perpendicular bisector will be a vertical line as the chord itself is a horizontal line, and vertical lines have their equation of the form π₯ equals constant. As the π₯-coordinate of the point that this line passes through is negative three, the equation of this line must be: π₯ equals negative three.

So now we have the equations of our two perpendicular bisectors, π₯ equals negative three and π¦ equals π₯. Remember, weβre looking to solve these simultaneously in order to find the coordinates of the centre of the circle. Well, theyβre quite straightforward equations to solve. We already know the value of π₯, and we know that π¦ is equal to π₯ which means that the point where these two lines meet must be the point with coordinates negative three, negative three. So this means that the centre of the circle whose equation weβre looking to find, is this point negative three, negative three.

Next, we need to find the radius of the circle which, remember, is the distance between the centre and any of the points on the circumference. Letβs look at the distance between the centre and point π΄. They both have the same π¦-coordinate of negative three, which means theyβre on a horizontal line. Therefore, the distance between these two points is just the difference between the π₯-coordinates, and the difference between negative three and negative four is one. This means that the radius of our circle is one length unit. Now you can confirm this by looking at the distance between the centre and point πΆ which are also on a horizontal line, or you could use the distance formula to calculate the distance between the centre and point π΅. But in any case, weβll get this value of one.

So now we know the centre and radius of the circle; we just need to find itβs equation. If we recall the standard form of the equation of a circle, the centre radius form, it tells us that if a circle has a centre with coordinates β, π and radius π, then its equation is: π₯ minus β all squared plus π¦ minus π all squared is equal to π squared. As we know both the centre and radius of our circle, we can just substitute the relevant values: β and π are both negative three, and π is one. So we have π₯ minus negative three all squared plus π¦ minus negative three all squared is equal to one squared.

Finally, we can just neaten this up a little. When you are subtracting negative three, that is equivalent to adding three. So we can simplify both brackets, and also one squared is equal to one. And so we have our answer to the problem.

The equation of the circle that passes through these two points is: π₯ plus three squared plus π¦ plus three squared is equal to one.