Find the centre of the circle through points 𝐴 three, one; 𝐵 one, two; and 𝐶 negative one, negative two.
So, the first thing I’ve done is I’ve drawn a sketch to help us understand what’s going on. So, we’ve got our circle. And we’ve got the three points that it goes through. Well, what we can do is we can call the centre of our circle 𝑀, so the middle of our circle. And we can give it coordinates 𝑎, 𝑏. Now, how is this helpful? Well, it’s helpful because we can say that the modulus, or magnitude, of 𝐴𝑀 is equal to the magnitude of 𝐶𝑀 is equal to the magnitude of 𝐵𝑀. And that’s because they’d all represent the radius.
Well, to find out these values, what we can use is the Pythagorean theorem. Which is 𝑎 squared plus 𝑏 squared equals 𝑐 squared, where we have a right-angle triangle, and 𝑎 and 𝑏 are the two shorter sides, and 𝑐 is our hypotenuse. To give us an example of this, what I’ve done is I’ve drawn the right-angle triangle that will be formed if we looked at 𝐶𝑀. Because we’re looking at coordinates, we can work out the two shorter sides. So, one would be 𝑎 minus negative one and one would be 𝑏 minus negative two cause it’s the change in our 𝑥-coordinates and the change in our 𝑦-coordinates. So therefore, we could use this to find out our 𝐶𝑀.
So, first of all, for our 𝐴𝑀, we’re gonna have three minus 𝑎 squared plus one minus 𝑏 all squared. And then, for our 𝐶𝑀, we’re gonna have 𝑎 plus one all squared plus 𝑏 plus two all squared. And then, for 𝐵𝑀, we’ve got one minus 𝑎 all squared plus two minus 𝑏 all squared. Okay, great. So, now, what should we do? Well, now, let’s distribute across our parentheses. So, when we distribute across our parentheses, we’re gonna get nine minus six 𝑎 plus 𝑎 squared plus one minus two 𝑏 plus 𝑏 squared equals 𝑎 squared plus two 𝑎 plus one plus 𝑏 squared plus four 𝑏 plus four. Which is then equal to 𝑎 squared minus two 𝑎 plus one plus 𝑏 squared minus four 𝑏 plus four.
So, the first thing we can see is what we can do is we can cancel 𝑎 square and 𝑏 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 𝐴𝑀 and our information for 𝐶𝑀. So, we get negative six 𝑎 minus two 𝑏 plus 10 equals negative two 𝑎 minus four 𝑏 plus five. So, then, what we can do is we can rearrange. And we do that by adding six 𝑎 and adding two 𝑏 to each side of the equation and also by subtracting five from each side of the equation.
So, we’re gonna get five is equal to four 𝑎 minus two 𝑏. And we’ve 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 𝐶𝑀 and 𝐵𝑀. So, when we do this, we’re gonna get two 𝑎 plus four 𝑏 plus five is equal to negative two 𝑎 minus four 𝑏 plus five. Well, we can see that in the first equation, we’ve got a term that’s four 𝑎. Well, then, in that case, what we’re gonna do is rearrange to see if we can get four 𝑎 on its own on one side of the equation in this second equation.
And we can because if we add on two 𝑎 to both sides of the equation and subtract four 𝑏 and subtract five, what we’re gonna get is four 𝑎 is equal to negative eight 𝑏. And that’s because two 𝑎 add two 𝑎 is four 𝑎. Negative four 𝑏 minus four 𝑏 is negative eight 𝑏. And then, five minus five is zero. Okay, great. So, now, what we’re gonna do is subtract our value for four 𝑎 into our first equation.
So, when we do that, we get five is equal to negative eight 𝑏 minus two 𝑏, which is gonna give us five is equal to negative 10𝑏. So, then, all we need to do is divide through by negative 10. And we’ll get our value for 𝑏, which is negative a half. Okay, great. So, now we found 𝑏, what we can do to just substitute this back into equation two to find 𝑎.
So, now, what we’re gonna get is four 𝑎 is equal to negative eight multiplied by negative a half. So, this is gonna give us four 𝑎 is equal to four. Then, we can divide through by four, which will give us 𝑎 is equal to one. So therefore, we found the coordinates for our centre of our circle cause we’ve got 𝑎 equals one and 𝑏 equals negative a half. So therefore, we can say that the coordinates for the centre of our circle are one, negative a half.