# Video: Finding the Point Which Lies on a Circle given Its Equation

Which of the following points lies on the circle 𝑦² + (𝑥 + 6)² = 144? [A] (11, 127) [B] (−6, 0) [C] (6, 0) [D] (6, 12)

03:06

### Video Transcript

Which of the following points lies on the circle 𝑦 squared plus 𝑥 plus six all squared equals 144? The options are 11, 127; negative six, zero; six, zero; or six, 12.

To answer this question, we can substitute each coordinate, that’s a pair of 𝑥- and 𝑦-values, into the equation of the circle and see if it does indeed satisfy the given equation. For example, for the first point, we substitute 11 for 𝑥 and 127 for 𝑦 into the left-hand side of the equation of the circle, giving 127 squared plus 11 plus six all squared.

Now the question is does this equal 144? If we have a calculator then, we can work it out. But if we don’t, then actually we can see straight away that it can’t equal 144 because 127 squared is gonna be a far bigger number than 144. In fact, 11 plus six is equal to 17. And if we add 127 and 17, we do get 144. But we certainly don’t get 144 if we add 127 squared and 17 squared. We’d get 144 if we forgot to square the values before adding them. So this coordinate does not satisfy the equation for circle. And therefore, this point does not lie on the circumference of the circle.

Let’s consider the next point, which has coordinates negative six, zero. Substituting negative six for 𝑥 and zero for 𝑦, we get zero squared plus negative six plus six squared. Negative six plus six is just equal to zero. So the second term becomes zero squared, which is zero. And again we’re adding zero squared to this. So overall, this gives zero. Zero is not equal to 144. So we can also rule out the second point.

Let’s consider the third point with coordinates six, zero. Substituting positive six for 𝑥 this time and zero for 𝑦 gives zero squared plus six plus six squared. Zero squared is zero. And six plus six is 12. 12 squared is equal to 144, which tells us that the point six, zero does lie on the circle with the given equation.

However, we’ve got one more point to check because it doesn’t say in the question that only one of these points lies on the circumference of the circle. For the final point, we substitute 12 for 𝑦 and six for 𝑥, giving 12 squared plus six plus six squared. Six plus six is 12. So we have 12 squared plus 12 squared again, which is 144 plus 144. This is therefore two lots of 144, 288, which is not equal to 144. So the final point doesn’t lie on the circle.

By substituting the coordinates of each point into the equation of the circle and checking whether or not they satisfy the equation, we found that the only one of the four points which lies on the circle with the given equation is the point six, zero.