Video: Pack 1 • Paper 2 • Question 22

Pack 1 • Paper 2 • Question 22

05:17

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.

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