### Video Transcript

A circle has equation 𝑥 squared
plus 𝑦 squared is equal to 45. Prove algebraically that the
straight line two 𝑥 minus 𝑦 is equal to 15 is a tangent to the circle.

Now, I’ll come back to what it
means for a line to be a tangent to a circle in a minute. But first, I’m just going to
rearrange the equation of the straight line slightly. I want to make 𝑦 the subject. So I’ll begin by adding 𝑦 to each
side, giving two 𝑥 is equal to 𝑦 plus 15. Next, I’ll subtract 15 from each
side, giving two 𝑥 minus 15 is equal to 𝑦. I’ve done that rearranging just so
that I can picture this line a little bit more accurately in the next stage. I now know that this is a straight
line with a positive gradient as its gradient is two.

Let’s think about what it means for
a straight line to be a tangent to a circle. Whenever a straight line and a
circle exist in the same coordinate plane, there’re three possibilities for the
relative position of the circle and the line. The first possibility is that the
line cuts through the circle. This means that the line meets the
circle in two places. There are two points of
intersection. The second possibility is that the
line just touches the circumference of the circle, but doesn’t pass inside, meaning
that there’s one point of intersection between the line and the circle. The third possibility is that the
line completely misses the circle and there are no points of intersection between
the two.

The second possibility would mean
that the line is a tangent to the circle, which is what we’ve been asked to prove in
this question. We, therefore, need to prove that
there is only one point of intersection between the line and the circle. To do so, we need to solve the
equation of the line and the equation of the circle simultaneously.

We’ve already rearranged the
equation of the line to give 𝑦 is equal to two 𝑥 minus 15. So we’re going to substitute this
into the equation of the circle. 𝑥 squared plus 𝑦 squared is equal
to 45. Therefore, becomes 𝑥 squared plus
two 𝑥 minus 15 all squared is equal to 45. This is an equation in 𝑥 only,
which we can solve.

We need to expand the bracket and
we must be careful here because a really common mistake to make when squaring a
bracket is to just square the two terms individually. Two 𝑥 minus 15 all squared is not
equal to two 𝑥 squared minus 15 squared. Instead, let’s write the repeated
bracket out twice and then we can use the FOIL method to expand.

Multiplying the first term in each
bracket together, two 𝑥 multiplied by two 𝑥 is equal to four 𝑥 squared. Multiplying the outer terms
together, we have two 𝑥 multiplied by negative 15 which is negative 30𝑥. Multiplying the inner terms
together, we have negative 15 multiplied by two 𝑥 which is also negative 30𝑥. Finally, multiplying the last term
in the two brackets together, we have negative 15 multiplied by negative 15 which is
225.

The equation therefore becomes 𝑥
squared plus four 𝑥 squared minus 60𝑥 plus 225 is equal to 45. And I’ve grouped the two lots of
negative 30𝑥 together to make negative 60𝑥. Now, 𝑥 squared plus four 𝑥
squared is equal to five 𝑥 squared. And at the same time, I’ve
subtracted 45 from both sides of the equation. So I now have five 𝑥 squared minus
60𝑥 plus 180 is equal to zero. I have a quadratic equation in 𝑥
which I’d like to solve. But I can make this simpler.

I noticed that all of the
coefficients are multiples of five. So I can simplify by dividing the
whole equation through by five. This gives a simple equation 𝑥
squared minus 12𝑥 plus 36 is equal to zero. I want to solve this equation for
𝑥. And to do so, I need to
factorize. You maybe able to spot straightaway
how this quadratic factorizes. But if not, let’s recall how to
factorize a quadratic equation.

As the coefficient of 𝑥 squared is
just one, the beginning terms in each bracket are each just 𝑥. To find the two numbers that go in
the bracket, I need to find two numbers that sum to the coefficient of 𝑥 — so
that’s negative 12 — and multiply it to the constant term, which is plus 36. The two numbers that do that are
negative six and negative six. So the two brackets are 𝑥 minus
six and 𝑥 minus six. Now, here’s the key point: these
two brackets are identical, which means if I were to solve this equation, I would
get 𝑥 minus six is equal to zero, meaning 𝑥 is equal to six. But there’s only one value for
𝑥.

If we think back to the three
situations we described, this means that there’s only one point of intersection
between the line and the circle. And therefore, the line is a
tangent to the circle. By solving the equation of the line
and the circle simultaneously, we’ve proved and algebraically that the straight line
is a tangent to the circle.