### Video Transcript

Find the centre of the circle through points π΄ three, one; π΅ one, two; and πΆ negative one, negative two.

So, the first thing Iβve done is Iβve drawn a sketch to help us understand whatβs going on. So, weβve got our circle. And weβve got the three points that it goes through. Well, what we can do is we can call the centre of our circle π, so the middle of our circle. And we can give it coordinates π, π. Now, how is this helpful? Well, itβs helpful because we can say that the modulus, or magnitude, of π΄π is equal to the magnitude of πΆπ is equal to the magnitude of π΅π. And thatβs because theyβd all represent the radius.

Well, to find out these values, what we can use is the Pythagorean theorem. Which is π squared plus π squared equals π squared, where we have a right-angle triangle, and π and π are the two shorter sides, and π is our hypotenuse. To give us an example of this, what Iβve done is Iβve drawn the right-angle triangle that will be formed if we looked at πΆπ. Because weβre looking at coordinates, we can work out the two shorter sides. So, one would be π minus negative one and one would be π minus negative two cause itβs the change in our π₯-coordinates and the change in our π¦-coordinates. So therefore, we could use this to find out our πΆπ.

So, first of all, for our π΄π, weβre gonna have three minus π squared plus one minus π all squared. And then, for our πΆπ, weβre gonna have π plus one all squared plus π plus two all squared. And then, for π΅π, weβve got one minus π all squared plus two minus π all squared. Okay, great. So, now, what should we do? Well, now, letβs distribute across our parentheses. So, when we distribute across our parentheses, weβre gonna get nine minus six π plus π squared plus one minus two π plus π squared equals π squared plus two π plus one plus π squared plus four π plus four. Which is then equal to π squared minus two π plus one plus π squared minus four π plus four.

So, the first thing we can see is what we can do is we can cancel π square and π squared throughout. And now, what we do is we can set up a pair of equations. So, the first equation we can set up is the equation which uses our information for π΄π and our information for πΆπ. So, we get negative six π minus two π plus 10 equals negative two π minus four π plus five. So, then, what we can do is we can rearrange. And we do that by adding six π and adding two π to each side of the equation and also by subtracting five from each side of the equation.

So, weβre gonna get five is equal to four π minus two π. And weβve called this equation one. So, now, we can take a look and see if we can form a second equation. Well, our second equation is gonna utilise the information for πΆπ and π΅π. So, when we do this, weβre gonna get two π plus four π plus five is equal to negative two π minus four π plus five. Well, we can see that in the first equation, weβve got a term thatβs four π. Well, then, in that case, what weβre gonna do is rearrange to see if we can get four π on its own on one side of the equation in this second equation.

And we can because if we add on two π to both sides of the equation and subtract four π and subtract five, what weβre gonna get is four π is equal to negative eight π. And thatβs because two π add two π is four π. Negative four π minus four π is negative eight π. And then, five minus five is zero. Okay, great. So, now, what weβre gonna do is subtract our value for four π into our first equation.

So, when we do that, we get five is equal to negative eight π minus two π, which is gonna give us five is equal to negative 10π. So, then, all we need to do is divide through by negative 10. And weβll get our value for π, which is negative a half. Okay, great. So, now we found π, what we can do to just substitute this back into equation two to find π.

So, now, what weβre gonna get is four π is equal to negative eight multiplied by negative a half. So, this is gonna give us four π is equal to four. Then, we can divide through by four, which will give us π is equal to one. So therefore, we found the coordinates for our centre of our circle cause weβve got π equals one and π equals negative a half. So therefore, we can say that the coordinates for the centre of our circle are one, negative a half.