Video: AQA GCSE Mathematics Higher Tier Pack 5 β€’ Paper 1 β€’ Question 28

A circle with the equation π‘₯Β² + 𝑦² = 20 is shown on the graph. The point 𝑆 lies on the circle and has a 𝑦-coordinate of 2. The tangent at 𝑆 intersects the π‘₯-axis at 𝑇. Find the coordinates of the point 𝑇.

07:00

Video Transcript

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

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

So next, what we’re 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, we’re not interested in the positive value of π‘₯. And that’s because if we take a look at where 𝑆 is, 𝑆 is to the left of the 𝑦-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 I’ve done is marked on the coordinates of our point 𝑆. And they are negative four, two. So now the reason that this is useful is because we can form a line. I’ve 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 𝑦 squared is equal to 20 or π‘₯ squared plus 𝑦 squared is equal to π‘Ÿ squared. And because we’ve got π‘₯ squared plus 𝑦 squared and there’s no additional numbers added or subtracted, then we know that the center of our circle is going to be zero, zero.

So now I’ve got this line β€” the radius β€” and I’ve got the tangent. But why are they useful? Well, it’s because of β€” as I mentioned before β€” the fact that they’re perpendicular to each other. And therefore, there’s a relationship between their gradients. But before we talk about that relationship and look at that: what we’re gonna do is work out the gradient of 𝑂𝑆. So I have called that π‘š cause π‘š is the gradient, so π‘š 𝑂𝑆.

And to help us work that out, we’ve got the gradient formula. And that is, the gradient π‘š is equal to 𝑦 two minus 𝑦 one over π‘₯ two minus π‘₯ one. And what this means is the change in 𝑦 divided by change in π‘₯. So then, what I’ve done is I’ve labelled our coordinates. So we’ve got π‘₯ one, 𝑦 one, π‘₯ two, 𝑦 two. I’m gonna substitute them into our formula. And when we do that, we’re gonna get zero minus two on the numerator. That’s cause it’s 𝑦 two minus 𝑦 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, we’re gonna get this equal to negative two over four, which we’ll 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 𝑆, so our radius of the circle.

As we mentioned, we’ve got a relationship between lines that are perpendicular. So if we got line one and line two and they’re perpendicular to each other, then if we multiply their gradients together, so π‘š one multiplied by π‘š two, it’s 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, I’ve shown π‘š two will be equal to negative one over π‘š 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, it’s gonna be positive because the gradient of 𝑂𝑆 was negative a half. So if you have a negative negative, then that’s just positive. And then if we find the reciprocal, well the reciprocal of one over two or a half is two over one. And that’s 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 𝑦 equals π‘šπ‘₯ plus 𝑐 and that’s because it’s a straight line. And we know π‘š is equal to two. So therefore, we’ve got 𝑦 equals two π‘₯ plus 𝑐. But we need to find out 𝑐. How are we going to do that?

So what we’re gonna do to find 𝑐 is sub in the π‘₯- and 𝑦-values for a point on the tangent. Well, we know a point on the tangent because we know 𝑆. The coordinates for 𝑆 are negative four, two. So therefore, we get two is equal to two multiplied by negative four plus 𝑐. So then, we’re gonna have two is equal to negative eight plus 𝑐. And then, when we add eight to each side of the equation, what we’re going to do is get 𝑐 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 𝑦 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 𝑇. Well, we know one of the coordinates of our point 𝑇. And that’s the second one, the 𝑦-coordinate. And we know that the 𝑦-coordinate is zero because we’re told that the tangent intersects the π‘₯-axis at 𝑇. And if it crosses the π‘₯-axis or intersects the π‘₯-axis, then at this point the 𝑦-coordinate is going to be zero. So what we need to do now is find out the π‘₯-coordinate of 𝑇. Well as we know that the 𝑦-coordinate is equal to zero, to find the π‘₯-coordinate of 𝑇, what we need to do is sub in 𝑦 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, we’re gonna get negative 10 is equal to two π‘₯. And then, what we’re gonna do is divide each side of the equation by two. And when we do that, I’ve 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 𝑇 are negative five, zero.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.