A circle with its centre at the origin passes through the point (0, 7).
What is the equation of this circle, and how did you work it out?
I dont really remeber could you walk me through the steps
It's easiest to start with the distance from the centre to the edge.
If the centre of your circle is at (0, 0) and it passes through the point (0, 7), what is the length of the radius?
That's right. Every point (x, y) on this circle is exactly 7 units away from the centre (0, 0).
If you imagine a right-angled triangle where the radius is the hypotenuse, and the other two sides are the horizontal distance x and the vertical distance y, what equation does Pythagoras' Theorem give you for this circle?
X squred plus b squared = c squared
You've got the right structure for Pythagoras' Theorem there.
In coordinate geometry, we use y for the vertical side. If the radius (the hypotenuse) is 7, what do you get when you write that equation using x, y, and 7?
X squared plus y squared = 7
You've got the x² + y² part spot on.
However, have a look at the right-hand side. Pythagoras' Theorem tells us that a² + b² = c². If the radius is 7, we need to square it for the equation.
What is the full equation for that circle? Once you've got that, try writing the equation for a circle with a radius of 10.
X squared plus y squared = 100
That's exactly it — you remembered to square the radius this time. For that first circle we looked at, the equation would be x² + y² = 49 because 7² = 49.
You've got the hang of the basic form x² + y² = r² for circles centred at the origin. Good work on clearing that up.
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