Video: AQA GCSE Mathematics Higher Tier Pack 2 β€’ Paper 1 β€’ Question 28

The diagram shows the circle π‘₯Β² + 𝑦² = 25 and the line 𝑦 = 3. Points 𝐴, 𝐡, and 𝐢 are points on the circumference of the circle. 𝐡 and 𝐢 lie on the line 𝑦 = 3. Point 𝐴 is on the 𝑦-axis. a) Show that the coordinates of 𝐴 are (0, βˆ’5). b) Find the coordinates of 𝐡 and 𝐢. c) Find the area of triangle 𝐴𝐡𝐢. d) Find the equation of the line 𝐴𝐢.

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Video Transcript

The diagram shows the circle π‘₯ squared plus 𝑦 squared equals 25 and the line 𝑦 is equal to three. Points 𝐴, 𝐡, and 𝐢 are points on the circumference of the circle. 𝐡 and 𝐢 lie on the line 𝑦 is equal to three. Point 𝐴 is on the 𝑦-axis. Part a) Show that the coordinates of 𝐴 are zero, negative five.

Recall the general equation of a circle with a centre at the origin β€” that’s zero, zero β€” and a radius of π‘Ÿ is π‘₯ squared plus 𝑦 squared equals π‘Ÿ squared. We can compare this to the equation of our circle. And we can see that our circle does indeed have a centre at the origin and that π‘Ÿ squared must be equal to 25.

We solve this equation for π‘Ÿ by square rooting both sides of the equation. The square root of 25 is plus or minus five. However, since π‘Ÿ is the radius of the circle which is a length, we’re going to disregard the negative value. And we can see that the radius of our circle is five units.

The line joining 𝑂 to 𝐴 is a line joining a point at the centre to a point on the circumference of the circle. By definition, that makes it the radius of the circle. And so, it’s five units. Since it’s five units directly below the centre of the circle which is at the origin, that means its coordinates are zero, negative five.

Part b) Find the coordinates of 𝐡 and 𝐢.

We are told that the chord that joins 𝐡 and 𝐢 has the equation 𝑦 is equal to three. This means the 𝑦-coordinates for 𝐡 and 𝐢 are both three. So we’re going to substitute 𝑦 equals three into the equation of our circle.

π‘₯ squared plus 𝑦 squared equals 25 becomes π‘₯ squared plus three squared equals 25. And of course, three squared is nine. It’s three multiplied by itself. So we have π‘₯ squared plus nine is equal to 25.

To solve this equation, we’ll subtract nine from both sides. And we get π‘₯ squared is equal to 16. Next, we’ll square root both sides of the equation. And when we square root 16, we get a positive and a negative solution.

This is because we know four multiplied by itself is 16. But also a negative multiplied by a negative is a positive. So negative four multiplied by negative four is also 16.

And our two possible values for the π‘₯-coordinate when the 𝑦-coordinate is three are positive four and negative four.

𝐡 lies on the left-hand side of the 𝑦-axis. So it must have a negative π‘₯-coordinate; it’s negative four, three. And 𝐢 lies on the right-hand side. So it’s got a positive π‘₯-coordinate; it’s four, three. 𝐡 is negative four, three and 𝐢 is four, three.

Now, it’s worth also noting we could have solved this a slightly different way. And that involved adding a right-angled triangle to our diagram.

The radius of the circle is five. So the hypotenuse, the longest side of this triangle, is five units. We saw that the 𝑦-coordinate was three. So the height of our triangle is three units. At this point, we could have used Pythagoras’s theorem to find the missing length.

Remember Pythagoras’s theorem says that the sum of the square of the smaller two sides is equal to the square of the longer side. That’s π‘₯ squared plus three squared equals five squared.

At this point, the process for solving this equation is identical to the one we went through before. And once again, we would have ended up with negative four, three and four, three for our two coordinates.

Part c) Find the area of triangle 𝐴𝐡𝐢.

Remember the formula for area of a triangle is a half π‘β„Ž, where 𝑏 is the base of the triangle and β„Ž is its perpendicular height. We can call the side 𝐡𝐢 the base of the triangle.

Since the π‘₯-coordinates for 𝐡 and 𝐢 are four and negative four, we can find the length of the line 𝐡𝐢 by finding the difference between these two numbers. That’s four minus negative four. Four minus negative four is four plus four which is eight. So the base of our triangle is eight units.

The height of the triangle is the line I’ve just added to the diagram. Let’s call the point where this line meets the base of the triangle 𝑃. And the height of the triangle is now the length of 𝑃 to 𝐴. We know the length of the line 𝑃 to 𝐴 is the length from 𝑂 to 𝑃 and 𝑂 to 𝐴 combined.

The line joining 𝑂 and 𝐴 is the radius of the circle. So it’s five units. And the length of the line joining 𝑂 to 𝑃 must be three units. It’s the line joining the origin to the point at which 𝑦 is equal to three. And we can see that the height of our triangle is eight units.

Substituting these values into our formula for the area of the triangle and we get a half multiplied by eight multiplied by eight which is 32. And the area of our triangle is 32 square units.

Part d) Find the equation of the line 𝐴𝐢.

To find the equation of the line joining 𝐴 to 𝐢, we recall the general formula for the equation of a straight line. It’s 𝑦 is equal to π‘šπ‘₯ plus 𝑐, where π‘š is the value of the gradient and 𝑐 is the value of the 𝑦-intercept of the line.

We’re going to need to find the gradient of the line 𝐴𝐢 then. Gradient is change in 𝑦 over change in π‘₯. That’s often written as 𝑦 two minus 𝑦 one over π‘₯ two minus π‘₯ one.

Don’t worry too much about these little numbers. It’s just reminding us to always subtract the coordinates in the same order. Let’s see what that looks like.

𝐴 is the point zero, negative five and 𝐢 is the point four, three. Let’s call the coordinate 𝐴 π‘₯ one and 𝑦 one and the coordinate 𝐢 π‘₯ two, 𝑦 two.

Then, 𝑦 two minus 𝑦 one is three minus negative five and π‘₯ two minus π‘₯ one is four minus zero. Three minus negative five is three plus five which is eight and four minus zero is four. Eight divided by four is two. So we’ve got a gradient of two.

Now, it’s important to realise that had we chosen π‘₯ one and π‘₯ two and 𝑦 one and 𝑦 two differently, we would have still got the same answer, as long as we always remember to make sure that π‘₯ one and 𝑦 one are part of the same coordinate.

So since our gradient π‘š is two, we can replace π‘š in the general equation with the number two. And 𝑦 is equal to two π‘₯ plus 𝑐. At this point, we can then substitute in either of the coordinates 𝐴 and 𝐢.

Let’s choose 𝐴. So we get negative five is equal to two multiplied by zero plus 𝑐. Remember zero is the π‘₯-value and negative five is the 𝑦-value. Two multiplied by zero is zero. So we get negative five equals zero plus 𝑐. Therefore, 𝑐 is equal to negative five.

And we can substitute 𝑐 back into the equation of a straight line. And we get 𝑦 is equal to two π‘₯ minus five.

Had we substituted in π‘₯ is equal to four and 𝑦 is equal to three from the coordinate labelled 𝐢, we would have still got the same equation: 𝑦 is equal to two π‘₯ minus five.

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