### 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 weβve done is drawn a sketch. So, weβve drawn a sketch of our circle. And itβs got the points negative six, negative nine; one, negative two; and π₯ sub one, zero on its a circumference. And weβve got that because our first point π₯ sub one, zero is because weβre told that the circle touches the π₯-axis. So therefore, we know that at that point, the π¦-value is going to be zero. And what Iβve also marked on is the centre of our circle π, π.

So, in this question, what weβre looking for is the equation of our circle. So, letβs take a look at the general form of any circle. And that is that π₯ minus π all squared plus π¦ minus π all squared is equal to π squared. Thatβs where the centre is π, π and the radius is π. And Iβve drawn π, our radius, onto our diagram here. Now, can we gather any information from the sketch with the information weβve been given?

Well, first of all, we have π₯ sub one is equal to π. And thatβs because we know that the π₯-axis acts as a tangent. So therefore, itβs gonna be perpendicular, or at right angles, to the radius. So therefore, that means that π₯ sub one and π are in-line. So, π₯ sub one is equal to the coordinate π. And we can also see that π squared is gonna be able to π squared. And thatβs 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 is gonna be equal to the magnitude of π. So therefore, π squared is equal to π squared.

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

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

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

So, now, what weβre 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, Iβm gonna look at the πs. And I want to have 18 πs in each of my equations. So, what Iβm gonna do is multiply equation one by 4.5. And when we do that, what weβre gonna get is 4.5π₯ sub one squared minus nine π₯ sub one plus 18π plus 22.5 equals zero.

So, then, what weβre gonna do is do equation three minus equation two. And we do this cause itβs going to eliminate our πs. So, what weβre 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, itβs in fact easier to divide through by seven. Thatβs 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 weβre 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 thatβs 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 youβd have negative three plus three, which would be zero. Or on the right-hand side of the parentheses, weβd 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 π¦-coordinate, which is gonna be π, is.

So therefore, to find out what our π-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 π plus five equals zero. Which would give nine plus six plus four π plus five equals zero. So, then, weβd have four π 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 π-values to be negative five. So therefore, we know that π is gonna be equal to five or, well, weβve got to find the other value.

And to find this other value, weβre gonna substitute in nine for π₯ sub one. And when we do this, weβre gonna get nine squared minus 18 plus four π plus five equals zero. So, thatβs 81 minus 18 plus four π plus five equals zero. So, this is gonna give us four π equals negative 68. So, all we have to do now is divide through by four. So, this will give us π 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.

Now, on our sketch, we obviously could only draw one of the circles. And that looks like weβve 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 squared, is going to be. And we know that π squared is equal to π squared. But therefore, π squared could be either negative five all squared, which is 25, or negative 17 all squared, which is 289. So, now, weβve 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 π¦ plus five all squared equals 25 or π₯ minus nine all squared plus π¦ plus 17 all squared equals 289. So, itβs worth noting that we need to be careful here with signs. Because if we look at the general equation of a circle, weβve got π₯ minus π and π¦ minus π. 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 weβve got π₯ plus three, π¦ plus five.