### Video Transcript

A circle with the equation π₯
squared plus π¦ squared equals 20 is shown on the graph. The point π lies on the circle and
has a π¦-coordinate of two. The tangent at π intersects the
π₯-axis at π. Find the coordinates of the point
π.

So the first thing weβre gonna do
is weβre gonna find the coordinates of the point π. But we already know the
π¦-coordinate of point π. And that is two because weβre told
that in the question. And we have an equation for the
circle. And that is that π₯ squared plus π¦
squared equals 20. So what weβre gonna do is use this
in conjunction with the π¦-coordinate to find the π₯-coordinate of the point π. So what weβre gonna do is
substitute in π¦ equals two. And when we do that, weβre gonna
get π₯ squared plus two squared equals 20. So therefore, weβre gonna get π₯
squared plus four is equal to 20.

So next, what weβre gonna do to get
the π₯ squared on its own is subtract four from each side of the equation. And when we do that, we get π₯
squared is equal to 16. So then, the next step to find π₯
is to take the square root of each side of the equation. And when we do that, we get π₯ is
equal to positive or negative four. However, weβre not interested in
the positive value of π₯. And thatβs because if we take a
look at where π is, π is to the left of the π¦-axis. So therefore, in this region, the
π₯-coordinates are all going to be negative. So therefore, we can say that π₯ is
equal to negative four.

So now what Iβve done is marked on
the coordinates of our point π. And they are negative four,
two. So now the reason that this is
useful is because we can form a line. Iβve done that with the dotted blue
line. This can be the radius of the
circle. And it goes from the point negative
four, two to the point zero, zero, so through the origin. And we also know that this is going
to be perpendicular or at right angles to our tangent.

And we know that the center of the
circle is zero, zero because our equation for the circle is in the form π₯ squared
plus π¦ squared is equal to 20 or π₯ squared plus π¦ squared is equal to π
squared. And because weβve got π₯ squared
plus π¦ squared and thereβs no additional numbers added or subtracted, then we know
that the center of our circle is going to be zero, zero.

So now Iβve got this line β the
radius β and Iβve got the tangent. But why are they useful? Well, itβs because of β as I
mentioned before β the fact that theyβre perpendicular to each other. And therefore, thereβs a
relationship between their gradients. But before we talk about that
relationship and look at that: what weβre gonna do is work out the gradient of
ππ. So I have called that π cause π
is the gradient, so π ππ.

And to help us work that out, weβve
got the gradient formula. And that is, the gradient π is
equal to π¦ two minus π¦ one over π₯ two minus π₯ one. And what this means is the change
in π¦ divided by change in π₯. So then, what Iβve done is Iβve
labelled our coordinates. So weβve got π₯ one, π¦ one, π₯
two, π¦ two. Iβm gonna substitute them into our
formula. And when we do that, weβre gonna
get zero minus two on the numerator. Thatβs cause itβs π¦ two minus π¦
one and then over zero minus negative four.

Well, we need to remember that if
we subtract a negative, this is same as adding. So therefore, weβre gonna get this
equal to negative two over four, which weβll simplify because we can divide the
numerator and denominator by two which will give us a gradient of negative a
half. Okay, so great, we found the
gradient of our π, so our radius of the circle.

As we mentioned, weβve got a
relationship between lines that are perpendicular. So if we got line one and line two
and theyβre perpendicular to each other, then if we multiply their gradients
together, so π one multiplied by π two, itβs gonna be equal to negative one. And if we rearrange this, we can
say that one of the gradients is gonna be equal to negative one over the other
gradient. So in this case, Iβve shown π two
will be equal to negative one over π one. And this is known as the negative
reciprocal. So therefore, we can use this to
work out what the gradient of our tangent is going to be.

So therefore, using this rule, the
gradient of the tangent is gonna be equal to- well, first of all, itβs gonna be
positive because the gradient of ππ was negative a half. So if you have a negative negative,
then thatβs just positive. And then if we find the reciprocal,
well the reciprocal of one over two or a half is two over one. And thatβs because to find the
reciprocal, we swap the numerator and the denominator. So therefore, we can say that the
gradient of the tangent is going to be two.

So now what we want to do is we
want to find the equation of the tangent. But how are we going to do
that? Well, we know the equation of the
tangent is gonna take the form π¦ equals ππ₯ plus π and thatβs because itβs a
straight line. And we know π is equal to two. So therefore, weβve got π¦ equals
two π₯ plus π. But we need to find out π. How are we going to do that?

So what weβre gonna do to find π
is sub in the π₯- and π¦-values for a point on the tangent. Well, we know a point on the
tangent because we know π. The coordinates for π are negative
four, two. So therefore, we get two is equal
to two multiplied by negative four plus π. So then, weβre gonna have two is
equal to negative eight plus π. And then, when we add eight to each
side of the equation, what weβre going to do is get π is equal to 10. So therefore, we can substitute
this back in. And we can say the equation of the
tangent is going to be π¦ equals two π₯ plus 10.

So now what we want to do is use
the equation of the tangent to find the coordinates of our point π. Well, we know one of the
coordinates of our point π. And thatβs the second one, the
π¦-coordinate. And we know that the π¦-coordinate
is zero because weβre told that the tangent intersects the π₯-axis at π. And if it crosses the π₯-axis or
intersects the π₯-axis, then at this point the π¦-coordinate is going to be
zero. So what we need to do now is find
out the π₯-coordinate of π. Well as we know that the
π¦-coordinate is equal to zero, to find the π₯-coordinate of π, what we need to do
is sub in π¦ is equal to zero to our equation.

And when we do that, we get zero is
equal to two π₯ plus 10. And then if we subtract 10 from
each side of the equation, weβre gonna get negative 10 is equal to two π₯. And then, what weβre gonna do is
divide each side of the equation by two. And when we do that, Iβve just
swapped the other way around just we have π₯ on the left, we get π₯ is equal to
negative five.

So therefore, we can say that the
coordinates of the point π are negative five, zero.