# Video: AQA GCSE Mathematics Higher Tier Pack 1 β’ Paper 1 β’ Question 20

AQA GCSE Mathematics Higher Tier Pack 1 β’ Paper 1 β’ Question 20

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### Video Transcript

A circle has the equation π₯ squared plus π¦ squared equals one-ninth. Circle the length of its radius. The options are one-ninth, one over 81, one-third, or one eighteenth.

The general equation of a circle with its center at the point π, π and a radius of π units is π₯ minus π squared plus π¦ minus π squared equals π squared. If the circle has its center at the origin, so both the values of π and π are zero, then this equation becomes π₯ squared plus π¦ squared equals π squared.

If we compare this with the equation of the circle weβve been given, then we can see that this circle is indeed a circle with its center at the origin. And comparing the right side of the two equations, we see that π squared is equal to one-ninth.

To find the value of π, the radius of the circle, we need to take the square root of each side of the equation, giving π equals the square root of one-ninth. Now, usually, when you solve an equation by square rooting, we need to take plus or minus the square root. So thatβs the positive and negative roots. But the radius of a circle is a length, so itβs positive, which means we only take the positive square root.

To work this out, we need to recall one of our rules of surds, which is that if we have the square root of a fraction, so the square root of π over π, then this is actually equal to the square root of the numerator over the square root of the denominator.

So the radius of our circle is equal to the square root of one over the square root of nine. The square root of one is just one as one times one is one. And the square root of nine is three as three times three is nine. So this simplifies to one over three or one-third.

So the radius of the circle with equation π₯ squared plus π¦ squared equals one-ninth is one-third.