Video: Finding the General Form of the Equation of a Circle

Find the general form of the equation of a circle that touches the π‘₯-axis and passes through the two points (βˆ’6, βˆ’9) and (1, βˆ’2).

08:19

Video Transcript

Find the general form of the equation of a circle that touches the π‘₯-axis and passes through the two points negative six, negative nine and one, negative two.

So, the first thing we’ve done is drawn a sketch. So, we’ve drawn a sketch of our circle. And it’s got the points negative six, negative nine; one, negative two; and π‘₯ sub one, zero on its a circumference. And we’ve got that because our first point π‘₯ sub one, zero is because we’re told that the circle touches the π‘₯-axis. So therefore, we know that at that point, the 𝑦-value is going to be zero. And what I’ve also marked on is the centre of our circle π‘Ž, 𝑏.

So, in this question, what we’re looking for is the equation of our circle. So, let’s take a look at the general form of any circle. And that is that π‘₯ minus π‘Ž all squared plus 𝑦 minus 𝑏 all squared is equal to π‘Ÿ squared. That’s where the centre is π‘Ž, 𝑏 and the radius is π‘Ÿ. And I’ve drawn π‘Ÿ, our radius, onto our diagram here. Now, can we gather any information from the sketch with the information we’ve been given?

Well, first of all, we have π‘₯ sub one is equal to π‘Ž. And that’s because we know that the π‘₯-axis acts as a tangent. So therefore, it’s gonna be perpendicular, or at right angles, to the radius. So therefore, that means that π‘₯ sub one and π‘Ž are in-line. So, π‘₯ sub one is equal to the coordinate π‘Ž. And we can also see that π‘Ÿ squared is gonna be able to 𝑏 squared. And that’s because the radius is the distance from the π‘₯-axis to the centre of the circle. So therefore, we know that the magnitude of π‘Ÿ is gonna be equal to the magnitude of 𝑏. So therefore, π‘Ÿ squared is equal to 𝑏 squared.

So, then, if we substitute this into the general form of the equation of a circle, we’re gonna get π‘₯ minus π‘₯ sub one all squared plus 𝑦 minus 𝑏 all squared equals 𝑏 squared. So, now, what we’re gonna do is substitute in a couple of our coordinate points. So, we can start with one, negative two. So, we’re gonna get one minus π‘₯ sub one all squared plus negative two minus 𝑏 all squared equal to 𝑏 squared. So, then, when we distribute across the parentheses, what we’re gonna get is one minus two π‘₯ sub one plus π‘₯ sub one squared plus four plus four 𝑏 plus 𝑏 squared equals 𝑏 squared. So, now, we can tidy this up.

First of all, on the left- and right-hand sides of the equation, we can subtract 𝑏 squared. So, these cancel. So, then, if we tidy up, what we’re gonna have is π‘₯ sub one all squared minus two π‘₯ sub one plus four 𝑏 plus five is equal to zero. And we can call this equation one. So, now, what we’re gonna do is substitute in our next point, which is negative six, negative nine.

And when we substitute that in, what we’re gonna get is negative six minus π‘₯ sub one all squared plus negative nine minus 𝑏 all squared is equal to 𝑏 squared. And when we distribute across the parentheses, what we’re gonna get is 36 plus 12π‘₯ sub one plus π‘₯ sub one all squared plus 81 plus 18𝑏 plus 𝑏 squared equals 𝑏 squared. Once again, we can subtract 𝑏 squared from each side of the equation. So, these cancel. So, then, when we tidy this up, we’re gonna get π‘₯ sub one squared plus 12π‘₯ sub one plus 18𝑏 plus 117 equals zero. And I’ve called this equation two.

So, now, what we’re gonna do is solve equations one and two simultaneously. And we do that by eliminating one of the variables. But first of all, we need to make sure we get the same amount. So, I’m gonna look at the 𝑏s. And I want to have 18 𝑏s in each of my equations. So, what I’m gonna do is multiply equation one by 4.5. And when we do that, what we’re gonna get is 4.5π‘₯ sub one squared minus nine π‘₯ sub one plus 18𝑏 plus 22.5 equals zero.

So, then, what we’re gonna do is do equation three minus equation two. And we do this cause it’s going to eliminate our 𝑏s. So, what we’re gonna get when we do this is 3.5π‘₯ sub one squared minus 21π‘₯ sub one minus 94.5 equals zero. Now, what we can do is multiply through by two. So, now, we have the quadratic seven π‘₯ sub one all squared minus 42π‘₯ sub one minus 189 equals zero. Well, we can actually go ahead and factor this to solve it. But we can see that before we do that, it’s in fact easier to divide through by seven. That’s cause each of the terms can be divided by seven. So, we get π‘₯ sub one all squared minus six π‘₯ sub one minus 27 equals zero.

So, now, what we want to do is factor this to enable us to solve it. And to factor it, what we need to do is find two factors that add together, or sum, to negative six, so the coefficient of our π‘₯ sub one. And then, they multiply together to give us negative 27. So therefore, our factors are gonna be positive three and negative nine. So, what we’re gonna have is π‘₯ sub one plus three multiplied by π‘₯ sub one minus nine is equal to zero. So therefore, π‘₯ sub one is gonna be equal to negative three or nine.

And that’s because to have a result of zero, it means that one of our parentheses has gotta be equal to zero. So, if we had π‘₯ sub one is negative three, then you’d have negative three plus three, which would be zero. Or on the right-hand side of the parentheses, we’d have nine minus nine, which would give us zero. So, great, this would work. So, we found our π‘₯ sub one. So, we can use this now to find out what our 𝑦-coordinate, which is gonna be 𝑏, is.

So therefore, to find out what our 𝑏-value is gonna be, we can substitute π‘₯ sub one equals negative three or nine into equation one. So, if we did negative three, we get negative three all squared minus two multiplied by negative three plus four 𝑏 plus five equals zero. Which would give nine plus six plus four 𝑏 plus five equals zero. So, then, we’d have four 𝑏 is equal to negative 20 because we subtract 20 from each side of the equation. Then, all we have to do is divide through by four. And we get one of our 𝑏-values to be negative five. So therefore, we know that 𝑏 is gonna be equal to five or, well, we’ve got to find the other value.

And to find this other value, we’re gonna substitute in nine for π‘₯ sub one. And when we do this, we’re gonna get nine squared minus 18 plus four 𝑏 plus five equals zero. So, that’s 81 minus 18 plus four 𝑏 plus five equals zero. So, this is gonna give us four 𝑏 equals negative 68. So, all we have to do now is divide through by four. So, this will give us 𝑏 is equal to negative 17. So therefore, we can say that the centre of our circle is going to be negative three, negative five or nine, negative 17. So, we could actually, in fact, have two centres of our circle. And we know that because we know that π‘₯ sub one is equal to π‘Ž.

Now, on our sketch, we obviously could only draw one of the circles. And that looks like we’ve drawn the circle that has centre negative three, negative five. Well, as we know, to be able to form our equation of our circle, what we need now is to find out what our radius, or what our π‘Ÿ squared, is going to be. And we know that π‘Ÿ squared is equal to 𝑏 squared. But therefore, π‘Ÿ squared could be either negative five all squared, which is 25, or negative 17 all squared, which is 289. So, now, we’ve got all the information we need to form the two possible equations of our circle.

So therefore, our final equation is going to be π‘₯ plus three all squared plus 𝑦 plus five all squared equals 25 or π‘₯ minus nine all squared plus 𝑦 plus 17 all squared equals 289. So, it’s worth noting that we need to be careful here with signs. Because if we look at the general equation of a circle, we’ve got π‘₯ minus π‘Ž and 𝑦 minus 𝑏. So therefore, we can see that the sign of our centre of our circle, well, the coordinate values, change sign when we have it in our final equation. So, for instance, we had a centre of negative three, negative five. So therefore, we we’ve got π‘₯ plus three, 𝑦 plus five.

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