### Video Transcript

Find the general form of the equation of a circle that is tangent to the π₯-axis and has centre negative 19, two.

We begin by recalling the general form of the equation of a circle whose centre is at π, π and whose radius is π. Itβs π₯ minus π all squared plus π¦ minus π all squared equals π squared. We do, in fact, know the coordinates of the centre of our circle. Theyβre negative 19, two. So, we can let π be equal to negative 19 and π be equal to two. But how do we work out the radius of our circle?

Well, letβs begin by drawing a little sketch. The circle might look a little something like this. Weβre told that the π₯-axis is a tangent to the circle, and thatβs really important. It means we can calculate the radius of the circle. Itβs the vertical distance between the centre, which is at negative 19, two, and the π₯-axis. It must be two units.

Substituting π, π, and π into our formula for the equation of a circle and we get π₯ minus negative 19 all squared plus π¦ minus two all squared equals two squared. Negative negative 19 is positive 19. So, our equation becomes π₯ plus 19 all squared plus π¦ minus two all squared equals two squared. Letβs distribute each set of parentheses.

We recall that when we square an expression, we multiply it by itself. So, π₯ plus 19 all squared is π₯ plus 19 times π₯ plus 19. Similarly, π¦ minus two all squared is π¦ minus two times π¦ minus two. And two squared is four. We then multiply the first term in each expression. π₯ times π₯ is π₯ squared. We multiply the outer terms. Thatβs 19π₯. We multiply the inner terms to get another 19π₯. And then, we multiply the last term in each expression. 19 squared is 361.

Weβll repeat this process for π¦ minus two times π¦ minus two. We multiply π¦ by π¦. Thatβs π¦ squared. We multiply π¦ by negative two to get negative two π¦. We then multiply negative two by π¦ to get another negative two π¦. And finally, we multiply negative two by negative two. That gives us four. Letβs neaten things up a little bit. 19π₯ plus 19π₯ is 38π₯. And negative two π¦ minus two π¦ is negative four π¦. Weβll subtract four from both sides of our equation. Writing our equation in descending powers of π₯ and π¦, we obtain π₯ squared plus π¦ squared plus 38π₯ minus four π¦ plus 361 equals zero. And we found the general form of the equation of the circle tangent to the π₯-axis with centre negative 19, two.

Now, there is a quick way that we can check this. The π₯-axis is a tangent to the circle. This means the circle must touch the π₯-axis at the point with coordinates negative 19, zero. So, we can substitute π₯ equals negative 19 and π¦ equals zero into our equation and make sure we do, indeed, get zero. That gives us negative 19 squared plus zero squared plus 38 times negative 19 minus four times zero plus 361, which is indeed zero.