Video: Finding the Equation of a Circle

Given 𝐴 (βˆ’2, βˆ’4) and 𝐡 (βˆ’10, 0), determine the equation of the circle for which 𝐴𝐡 is a diameter.

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Video Transcript

Given the point 𝐴 is negative two, negative four and the point 𝐡 is negative 10, zero, determine the equation of the circle for which 𝐴𝐡 is the diameter.

So the general form for the equation of a circle is π‘₯ minus β„Ž all squared plus 𝑦 minus π‘˜ all squared equals π‘Ÿ squared. And this is where the center is β„Ž, π‘˜ and π‘Ÿ is the radius. So β„Ž is our π‘₯-coordinate, and π‘˜ is the 𝑦-coordinate at the center. And like we said, π‘Ÿ is the radius. Okay, great! So we know that. So what are we gonna do to solve the problem?

Well the first thing we wanna do is actually find the radius. And we can do that by calculating 𝐴𝐡 because we know that that’s the diameter, and then we can halve it to give us the radius. And the way we’re gonna do that is actually by using the distance formula, which is that 𝑑 is equal to the square root of π‘₯ two minus π‘₯ one all squared plus 𝑦 two minus 𝑦 one all squared. And we get this formula actually from the Pythagorean theorem. And that’s because 𝑑, our distance and in this case it’s gonna be 𝐴𝐡, is the hypotenuse and actually the two other sides can be found with π‘₯ two minus π‘₯ one and 𝑦 two minus 𝑦 one.

And we’ll show that exactly how it works when we solve this problem. But what I’ve done to actually help highlight this is I’ve actually drawn a sketch. So we’ve got our point 𝐴 negative two, negative four and our point 𝐡 negative 10, zero. So now what I’ve done is actually constructed a right-angled triangle just showing the difference between our π‘₯-coordinates and 𝑦-coordinates. So the difference between our π‘₯-coordinates would be eight, and the difference between our 𝑦-coordinates would be four.

And therefore, we could actually use the Pythagorean theorem to find the length 𝐴𝐡. So what I’ve done is I’ve actually labeled our points. So we’ve got the coordinates labeled with π‘₯ one, 𝑦 one and π‘₯ two, 𝑦 two because it’s gonna help us when we put them into our formula for the distance, in this case 𝐴𝐡. So therefore, we can say that 𝐴𝐡 is gonna be equal to the square root of negative 10 minus negative two, then this is all squared, plus zero minus negative four. And then that’s all squared. That’s because we had our π‘₯ two minus our π‘₯ one and our 𝑦 two minus our 𝑦 one.

And this is gonna give us that 𝐴𝐡 is equal to the square root of negative eight all squared. And that’s because negative 10 minus negative two is gonna be negative 10 plus two, so it gives us negative eight, and then plus four squared. So therefore, we can say that 𝐴𝐡 is gonna be equal to root 80. And that’s because negative eight all squared is 64 because a negative multiplied by a negative is positive, and four squared is 16. 64 plus 16 is 80. So we got 𝐴𝐡 is equal to the root 80.

Well now if we’ve actually simplify the surd, so we simplify root 80, this can be written as root 16 multiplied by root five. That’s because 16 is the highest square number that actually goes into 80, which just gives us four root five. So great! We now know the length 𝐴𝐡. So therefore, we can say that the radius of our circle is going to be equal to two root five. That’s because 𝐴𝐡 was our diameter. So therefore, the radius is half of this. So if you divide four root five by two, you get two root five.

Okay, great! So now we’ve got the radius. What we need to do is actually find the center Point. So in order to actually find out what the center of our line is going to be, so the center of 𝐴𝐡, because 𝐴𝐡 is a diameter, the center of it is actually gonna be the center of the circle. Then we use this formula for the midpoint, which is π‘₯ one plus π‘₯ two divided by two gives us the π‘₯-coordinate and 𝑦 one plus 𝑦 two over two give us the 𝑦- coordinate.

So again I’ve actually got the label coordinates π‘₯ one, 𝑦 one π‘₯ two, 𝑦 two. So we can put them into our formula to find our midpoint cause basically what we’re trying to do with midpoint is find the difference between our π‘₯-values and the difference between our 𝑦-values. And that will be our π‘₯- and 𝑦-values. So in this case, it’s gonna be our center of our circle. So our midpoint is gonna be equal to negative two plus negative 10 over two for our π‘₯-coordinate and negative four plus zero over two for our 𝑦-coordinate.

So therefore, the midpoint is gonna be equal to negative six, negative two. And that’s because we had negative two plus negative 10, or that’s a plus negative 10 turns into just negative two minus 10, which gives us minus 12. Minus 12 over two is minus six or negative six. Then we’ve got negative four plus zero, which is just negative four, negative four over two, which gives us negative two. Great! So that’s our midpoint found.

So now what we need to do is actually look at the equation of circle and see how we’re gonna use the information we’ve got to actually give us an equation of our circle. Well we know that π‘Ÿ is equal to two root five. However, for the equation of circle, we want to know what π‘Ÿ squared is. So we need to do is two root five all squared. Well two root five all squared is the same as two squared multiplied by root five squared. So we can say that π‘Ÿ squared is gonna be equal to four multiplied by five. And that’s because two squared is equal to four and root five multiplied by root five would give us root 25, which is just five.

So therefore, we can say that π‘Ÿ squared is gonna be equal to 20. Okay, great! So we can put this into the equation of the circle. So when we substitute our center point and our radius for our circle into the equation that we’ve got for the circle, we can say that π‘₯ minus negative six all squared plus 𝑦 minus negative two all squared is equal to 20. And we got the negative six because it was our π‘₯-coordinate of our midpoint. We got the negative two cause this was the 𝑦- coordinate of our midpoint.

So therefore, if we actually tidy this up, and we can tide this up because we got π‘₯ minus negative six, well if you have minus and negative it turns into a positive. So we’ve got π‘₯ plus six all squared plus 𝑦 plus two all squared equals 20. So therefore, we can say that given that 𝐴 is negative two, negative four, point 𝐡 is negative 10, zero, the equation of the circle for which 𝐴𝐡 is a diameter is π‘₯ plus six all squared plus 𝑦 plus two all squared equals 20.

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