Video Transcript
Find the general equation of the circle through the origin that also passes through 12, zero and zero, 16.
So the first thing I’ve done in this question is I’ve drawn a sketch. And this sketch shows the circle that we’re talking about. So we’ve got a circle that goes through the origin, so the point zero, zero. But it also passes through 12, zero and zero, 16. Well, if we’re trying to find out the equation of our circle, then there’re a couple of things that we’re gonna need. We’re gonna need the radius, and we’re gonna need the center of our circle. So what I’ve done is I’ve drawn a line. And this line is from the two points zero, 16 and 12, zero, and it joins them.
Well, if I call each of these points 𝐴 and 𝐵, then I can say that 𝐴𝐵 is going to be a diameter. And we know that it’s a diameter because it subtends 90-degree angle at the origin, which I’ve shown here. And we know it does that because we’ve got the origin is the zero, zero point. And then, 𝐴 and 𝐵 are both on our axes. So we know that they are in fact perpendicular to each other or at right angles to each other. Okay, great.
Well, we said that we needed to know what the radius and the center of our circle are. But why is that? Well, if we look at the general equation of a circle, it’s 𝑥 minus 𝑎 all squared plus 𝑦 minus 𝑏 all squared is equal to 𝑟 squared. And this is where the center is 𝑎, 𝑏 and the radius is equal to 𝑟. So first of all, let’s work out the radius. And we can do this by working out how long 𝐴𝐵 is, so how long the line 𝐴𝐵 is, and then halving it because the radius is half the diameter.
Well, as we can already see, we’ve got a right triangle, where we’ve got the hypotenuse, which is 𝐴𝐵. And then we’ve got one side of 16 and the other side of 12. So therefore, 𝐴𝐵 is going to be equal to the square root of 16 squared plus 12 squared. So therefore, 𝐴𝐵 is gonna be equal to the square root of 400. So 𝐴𝐵 is gonna give us 20. That’s cause the square root of 400 is 20. So therefore, as we know that 𝐴𝐵 is the diameter, then 𝑟 is gonna be half of this. So 𝑟, our radius, is gonna be equal to 10.
So now we’ve got the radius, what we need to find so that we can form the equation of our circle is the center of the circle. Well, to help us find the center of the circle, what we have is a little formula. And this formula is for the midpoint of a straight line. And that’s because our line 𝐴𝐵 is a straight line. And because it’s the diameter, it means it must go through the center of our circle. So therefore, the center of our circle is gonna be halfway through this line. So we’ve got 𝑥 one plus 𝑥 two over two is for the 𝑥-coordinate and 𝑦 one plus 𝑦 two over two is for our 𝑦-coordinate.
Well, what I’ve done is I’ve labeled our points 𝑥 one, 𝑦 one; 𝑥 two, 𝑦 two. So therefore, using this, we’re gonna have the center is equal to 12 plus zero over two. So that’s our 𝑥 one plus 𝑥 two over two. And then, for our 𝑦-coordinate is zero plus 16 over two. So therefore, our center of our circle is gonna be the coordinates six, eight. That’s cause we had 12 over two for the 𝑥-coordinate and 16 over two for the 𝑦-coordinate. Great, so now we have all the pieces we need to put this back together to form our equation of the circle. And then what we’re gonna have is 𝑥 minus six all squared plus 𝑦 minus eight all squared is equal to 10 squared. That’s cause our 𝑎 was equal to six, our 𝑏 was equal to eight, and our 𝑟 is equal to 10.
So then, when we distribute across our parentheses, we’re gonna get 𝑥 squared minus 12𝑥 plus 36. And that’s cause 𝑥 minus six all squared is the same as 𝑥 minus six multiplied by 𝑥 minus six. So then, we’ve got plus 𝑦 squared minus 16𝑦 plus 64, and then this is equal to 100. Well then, if we subtract 36 and 64 from the left-hand side, that’s gonna be 100. And therefore, if we subtract 100 from the right-hand side, we’re gonna get zero on the right-hand side.
So therefore, we’re gonna get our final general equation of our circle, which is gonna be 𝑥 squared plus 𝑦 squared minus 12𝑥 minus 16𝑦 is equal to zero.
